Mixing
Random variable defined as A with 50% chance and B with 50% chance
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up vote
7
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Note: this is a homework problem so please don't give me the whole answer!
I have two variables, A and B, with normal distributions (means and variances are known). Suppose C is defined as A with 50% chance and B with 50% chance. How would I go about proving whether C is also normally distributed, and if so, what its mean and variance are?
I'm not sure how to combine the PDFs of A and B this way, but ideally if someone can point me in the right direction, my plan of attack is to derive the PDF of C and show whether it is or isn't a variation of the normal PDF.
self-study random-variable gaussian-mixture finite-mixture-model
edited Aug 27 at 20:01
whuber♦
195k31417776
asked Aug 27 at 19:11
Bluefire
1455
 |Â
show 1 more comment
up vote
7
down vote
favorite
Note: this is a homework problem so please don't give me the whole answer!
I have two variables, A and B, with normal distributions (means and variances are known). Suppose C is defined as A with 50% chance and B with 50% chance. How would I go about proving whether C is also normally distributed, and if so, what its mean and variance are?
I'm not sure how to combine the PDFs of A and B this way, but ideally if someone can point me in the right direction, my plan of attack is to derive the PDF of C and show whether it is or isn't a variation of the normal PDF.
self-study random-variable gaussian-mixture finite-mixture-model
edited Aug 27 at 20:01
whuber♦
195k31417776
asked Aug 27 at 19:11
Bluefire
1455
2
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
6
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
4
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
3
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
5
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51
 |Â
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Note: this is a homework problem so please don't give me the whole answer!
I have two variables, A and B, with normal distributions (means and variances are known). Suppose C is defined as A with 50% chance and B with 50% chance. How would I go about proving whether C is also normally distributed, and if so, what its mean and variance are?
I'm not sure how to combine the PDFs of A and B this way, but ideally if someone can point me in the right direction, my plan of attack is to derive the PDF of C and show whether it is or isn't a variation of the normal PDF.
self-study random-variable gaussian-mixture finite-mixture-model
edited Aug 27 at 20:01
whuber♦
195k31417776
asked Aug 27 at 19:11
Bluefire
1455
Note: this is a homework problem so please don't give me the whole answer!
I have two variables, A and B, with normal distributions (means and variances are known). Suppose C is defined as A with 50% chance and B with 50% chance. How would I go about proving whether C is also normally distributed, and if so, what its mean and variance are?
I'm not sure how to combine the PDFs of A and B this way, but ideally if someone can point me in the right direction, my plan of attack is to derive the PDF of C and show whether it is or isn't a variation of the normal PDF.
self-study random-variable gaussian-mixture finite-mixture-model
edited Aug 27 at 20:01
whuber♦
195k31417776
asked Aug 27 at 19:11
Bluefire
1455
edited Aug 27 at 20:01
whuber♦
195k31417776
edited Aug 27 at 20:01
whuber♦
195k31417776
edited Aug 27 at 20:01
whuber♦
195k31417776
195k31417776
asked Aug 27 at 19:11
Bluefire
1455
asked Aug 27 at 19:11
Bluefire
1455
asked Aug 27 at 19:11
Bluefire
1455
1455
2
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
6
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
4
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
3
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
5
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51
 |Â
show 1 more comment
2
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
6
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
4
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
3
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
5
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51
2
2
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
6
6
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
4
4
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
3
3
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
5
5
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51
 |Â
show 1 more comment
5 Answers
5
active
oldest
votes
up vote
7
down vote
accepted
Hopefully it's clear to you that C isn't guaranteed to be normal. However, part of your question was how to write down its PDF.
@BallpointBen gave you a hint. If that's not enough, here are some more spoilers...
Note that C can be written as:
$$C = T cdot A + (1-T) cdot B$$
for a Bernoulli random $T$ with $P(T=0)=P(T=1)=1/2$ with $T$ independent of $(A,B)$. This is more or less the standard mathematical translation of the English statement "C is A with 50% chance and B with 50% chance".
Now, determining the PDF of C directly from this seems hard, but you can make progress by writing down the distribution function $F_C$ of C. You can partition the event $C leq X$ into two subevents (depending on the value of $T$) to write:
$$ F_C(x) = P(C leq x) =
P(T = 0 text and C leq x) + P(T = 1text and C leq x) $$
and note that by the definition of C and the independence of T and B, you have:
$$P(T=0text and C leq x) = P(T=0text and Bleq x) = frac12P(Bleq x) = frac12 F_B(x)$$
You should be able to use a similar result in the $T=1$ case to write $F_C$ in terms of $F_A$ and $F_B$. To get the PDF of C, just differentiate $F_C$ with respect to x.
answered Aug 27 at 20:58
K. A. Buhr
1861
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
add a comment |Â
up vote
8
down vote
Simulation of a random 50-50 mixture of $mathsfNorm(mu=90, sigma=2)$ and
$mathsfNorm(mu=100, sigma=2)$ is illustrated below. Simulation in R.
set.seed(827); m = 10^6
x1 = rnorm(m, 100, 2); x2 = rnorm(m, 90, 2)
p = rbinom(m, 1, .5)
x = x1; x[p==1] = x2[p==1]
hist(x, prob=T, col="skyblue2", main="Random 50-50 Mixture of NORM(90,2) and NORM(100,2)")
curve(.5*(dnorm(x, 100, 2) + dnorm(x, 90, 2)), add=T, col="red", lwd=2)
answered Aug 27 at 19:34
BruceET
2,610417
add a comment |Â
up vote
7
down vote
One way you could work on that is to analyze it as the variance tends to 0. This way you would get a Bernoulli-like distribution, which is (clearly) not a normal distribution.
answered Aug 27 at 19:34
André Costa
964
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
add a comment |Â
up vote
1
down vote
If the PDFs of $A$, $B$ and $C$ are $P_A(x)$, $P_B(x)$ and $P_C(x)$ respectivelly, and $alpha$ is the chance that $C$ is defined as $A$ (and $1-alpha$ is the chance that $C$ is defined as $B$), then
$$P_C(x) = alpha P_A(x) + (1-alpha) P_B(x).$$
answered Aug 28 at 14:29
HelloGoodbye
22827
2
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
add a comment |Â
up vote
1
down vote
This is the kind of problem where it is very helpful to use the concept of the CDF, the cumulative probability distribution function, of random variables, that totally unnecessary concept that professors drag in just to confuse students who are happy to just use pdfs.
By definition, the value of the CDF $F_X(alpha)$ of a random variable $X$ equals the probability that $X$ is no larger than the real number $alpha$, that is,
$$F_X(alpha) = PX leq alpha, ~-infty < alpha < infty.$$
Now, the law of total probability tells us that if $X$ is equally likely to be the same as a random variable $A$ or a random variable $B$, then
$$PX leq alpha = frac 12 PA leq alpha + frac 12 PB leq alpha,$$
or, in other words,
$$F_X(alpha} = frac 12 F_A(alpha} + frac 12 F_B(alpha}.$$
Remembering how your professor boringly nattered on and on about how for continuous random variables the pdf is the derivative of the CDF, we get that
$$f_X(alpha} = frac 12 f_A(alpha} + frac 12 f_B(alpha} tag1$$
which answers one of your questions. For the special case of normal random variables $A$ and $B$, can you figure out whether $(1)$ gives a normal density for $X$ or not? If you are familiar with notions such as
$$E[X] = int_-infty^infty alpha f_X(alpha} , mathrm dalpha,
tag2$$
can you figure out, by substituting the right side of $(1)$ for the $f_X(alpha)$ in $(2)$ and thinking about the expression, what $E[X]$ is in terms of $E[A]$ and $E[B]$?
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Hopefully it's clear to you that C isn't guaranteed to be normal. However, part of your question was how to write down its PDF.
@BallpointBen gave you a hint. If that's not enough, here are some more spoilers...
Note that C can be written as:
$$C = T cdot A + (1-T) cdot B$$
for a Bernoulli random $T$ with $P(T=0)=P(T=1)=1/2$ with $T$ independent of $(A,B)$. This is more or less the standard mathematical translation of the English statement "C is A with 50% chance and B with 50% chance".
Now, determining the PDF of C directly from this seems hard, but you can make progress by writing down the distribution function $F_C$ of C. You can partition the event $C leq X$ into two subevents (depending on the value of $T$) to write:
$$ F_C(x) = P(C leq x) =
P(T = 0 text and C leq x) + P(T = 1text and C leq x) $$
and note that by the definition of C and the independence of T and B, you have:
$$P(T=0text and C leq x) = P(T=0text and Bleq x) = frac12P(Bleq x) = frac12 F_B(x)$$
You should be able to use a similar result in the $T=1$ case to write $F_C$ in terms of $F_A$ and $F_B$. To get the PDF of C, just differentiate $F_C$ with respect to x.
answered Aug 27 at 20:58
K. A. Buhr
1861
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
add a comment |Â
up vote
7
down vote
accepted
Hopefully it's clear to you that C isn't guaranteed to be normal. However, part of your question was how to write down its PDF.
@BallpointBen gave you a hint. If that's not enough, here are some more spoilers...
Note that C can be written as:
$$C = T cdot A + (1-T) cdot B$$
for a Bernoulli random $T$ with $P(T=0)=P(T=1)=1/2$ with $T$ independent of $(A,B)$. This is more or less the standard mathematical translation of the English statement "C is A with 50% chance and B with 50% chance".
Now, determining the PDF of C directly from this seems hard, but you can make progress by writing down the distribution function $F_C$ of C. You can partition the event $C leq X$ into two subevents (depending on the value of $T$) to write:
$$ F_C(x) = P(C leq x) =
P(T = 0 text and C leq x) + P(T = 1text and C leq x) $$
and note that by the definition of C and the independence of T and B, you have:
$$P(T=0text and C leq x) = P(T=0text and Bleq x) = frac12P(Bleq x) = frac12 F_B(x)$$
You should be able to use a similar result in the $T=1$ case to write $F_C$ in terms of $F_A$ and $F_B$. To get the PDF of C, just differentiate $F_C$ with respect to x.
answered Aug 27 at 20:58
K. A. Buhr
1861
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Hopefully it's clear to you that C isn't guaranteed to be normal. However, part of your question was how to write down its PDF.
@BallpointBen gave you a hint. If that's not enough, here are some more spoilers...
Note that C can be written as:
$$C = T cdot A + (1-T) cdot B$$
for a Bernoulli random $T$ with $P(T=0)=P(T=1)=1/2$ with $T$ independent of $(A,B)$. This is more or less the standard mathematical translation of the English statement "C is A with 50% chance and B with 50% chance".
Now, determining the PDF of C directly from this seems hard, but you can make progress by writing down the distribution function $F_C$ of C. You can partition the event $C leq X$ into two subevents (depending on the value of $T$) to write:
$$ F_C(x) = P(C leq x) =
P(T = 0 text and C leq x) + P(T = 1text and C leq x) $$
and note that by the definition of C and the independence of T and B, you have:
$$P(T=0text and C leq x) = P(T=0text and Bleq x) = frac12P(Bleq x) = frac12 F_B(x)$$
You should be able to use a similar result in the $T=1$ case to write $F_C$ in terms of $F_A$ and $F_B$. To get the PDF of C, just differentiate $F_C$ with respect to x.
answered Aug 27 at 20:58
K. A. Buhr
1861
Hopefully it's clear to you that C isn't guaranteed to be normal. However, part of your question was how to write down its PDF.
@BallpointBen gave you a hint. If that's not enough, here are some more spoilers...
Note that C can be written as:
$$C = T cdot A + (1-T) cdot B$$
for a Bernoulli random $T$ with $P(T=0)=P(T=1)=1/2$ with $T$ independent of $(A,B)$. This is more or less the standard mathematical translation of the English statement "C is A with 50% chance and B with 50% chance".
Now, determining the PDF of C directly from this seems hard, but you can make progress by writing down the distribution function $F_C$ of C. You can partition the event $C leq X$ into two subevents (depending on the value of $T$) to write:
$$ F_C(x) = P(C leq x) =
P(T = 0 text and C leq x) + P(T = 1text and C leq x) $$
and note that by the definition of C and the independence of T and B, you have:
$$P(T=0text and C leq x) = P(T=0text and Bleq x) = frac12P(Bleq x) = frac12 F_B(x)$$
You should be able to use a similar result in the $T=1$ case to write $F_C$ in terms of $F_A$ and $F_B$. To get the PDF of C, just differentiate $F_C$ with respect to x.
answered Aug 27 at 20:58
K. A. Buhr
1861
answered Aug 27 at 20:58
K. A. Buhr
1861
answered Aug 27 at 20:58
K. A. Buhr
1861
answered Aug 27 at 20:58
K. A. Buhr
1861
1861
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
add a comment |Â
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
1
1
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
Notably, it follows from this answer that $C$ could be normal, e.g. when $A, B$ are identically distributed.
– Mees de Vries
Aug 28 at 10:13
add a comment |Â
up vote
8
down vote
Simulation of a random 50-50 mixture of $mathsfNorm(mu=90, sigma=2)$ and
$mathsfNorm(mu=100, sigma=2)$ is illustrated below. Simulation in R.
set.seed(827); m = 10^6
x1 = rnorm(m, 100, 2); x2 = rnorm(m, 90, 2)
p = rbinom(m, 1, .5)
x = x1; x[p==1] = x2[p==1]
hist(x, prob=T, col="skyblue2", main="Random 50-50 Mixture of NORM(90,2) and NORM(100,2)")
curve(.5*(dnorm(x, 100, 2) + dnorm(x, 90, 2)), add=T, col="red", lwd=2)
answered Aug 27 at 19:34
BruceET
2,610417
add a comment |Â
up vote
8
down vote
Simulation of a random 50-50 mixture of $mathsfNorm(mu=90, sigma=2)$ and
$mathsfNorm(mu=100, sigma=2)$ is illustrated below. Simulation in R.
set.seed(827); m = 10^6
x1 = rnorm(m, 100, 2); x2 = rnorm(m, 90, 2)
p = rbinom(m, 1, .5)
x = x1; x[p==1] = x2[p==1]
hist(x, prob=T, col="skyblue2", main="Random 50-50 Mixture of NORM(90,2) and NORM(100,2)")
curve(.5*(dnorm(x, 100, 2) + dnorm(x, 90, 2)), add=T, col="red", lwd=2)
answered Aug 27 at 19:34
BruceET
2,610417
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Simulation of a random 50-50 mixture of $mathsfNorm(mu=90, sigma=2)$ and
$mathsfNorm(mu=100, sigma=2)$ is illustrated below. Simulation in R.
set.seed(827); m = 10^6
x1 = rnorm(m, 100, 2); x2 = rnorm(m, 90, 2)
p = rbinom(m, 1, .5)
x = x1; x[p==1] = x2[p==1]
hist(x, prob=T, col="skyblue2", main="Random 50-50 Mixture of NORM(90,2) and NORM(100,2)")
curve(.5*(dnorm(x, 100, 2) + dnorm(x, 90, 2)), add=T, col="red", lwd=2)
answered Aug 27 at 19:34
BruceET
2,610417
Simulation of a random 50-50 mixture of $mathsfNorm(mu=90, sigma=2)$ and
$mathsfNorm(mu=100, sigma=2)$ is illustrated below. Simulation in R.
set.seed(827); m = 10^6
x1 = rnorm(m, 100, 2); x2 = rnorm(m, 90, 2)
p = rbinom(m, 1, .5)
x = x1; x[p==1] = x2[p==1]
hist(x, prob=T, col="skyblue2", main="Random 50-50 Mixture of NORM(90,2) and NORM(100,2)")
curve(.5*(dnorm(x, 100, 2) + dnorm(x, 90, 2)), add=T, col="red", lwd=2)
answered Aug 27 at 19:34
BruceET
2,610417
edited Aug 27 at 19:46
answered Aug 27 at 19:34
BruceET
2,610417
answered Aug 27 at 19:34
BruceET
2,610417
answered Aug 27 at 19:34
BruceET
2,610417
2,610417
add a comment |Â
add a comment |Â
up vote
7
down vote
One way you could work on that is to analyze it as the variance tends to 0. This way you would get a Bernoulli-like distribution, which is (clearly) not a normal distribution.
answered Aug 27 at 19:34
André Costa
964
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
add a comment |Â
up vote
7
down vote
One way you could work on that is to analyze it as the variance tends to 0. This way you would get a Bernoulli-like distribution, which is (clearly) not a normal distribution.
answered Aug 27 at 19:34
André Costa
964
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
add a comment |Â
up vote
7
down vote
up vote
7
down vote
One way you could work on that is to analyze it as the variance tends to 0. This way you would get a Bernoulli-like distribution, which is (clearly) not a normal distribution.
answered Aug 27 at 19:34
André Costa
964
One way you could work on that is to analyze it as the variance tends to 0. This way you would get a Bernoulli-like distribution, which is (clearly) not a normal distribution.
answered Aug 27 at 19:34
André Costa
964
answered Aug 27 at 19:34
André Costa
964
answered Aug 27 at 19:34
André Costa
964
answered Aug 27 at 19:34
André Costa
964
964
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
add a comment |Â
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
1
1
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
I didn't post is as a comment because I don't have enough reputation
– André Costa
Aug 27 at 19:35
1
1
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
Nevertheless, a good suggestion. (+1)
– BruceET
Aug 27 at 19:43
add a comment |Â
up vote
1
down vote
If the PDFs of $A$, $B$ and $C$ are $P_A(x)$, $P_B(x)$ and $P_C(x)$ respectivelly, and $alpha$ is the chance that $C$ is defined as $A$ (and $1-alpha$ is the chance that $C$ is defined as $B$), then
$$P_C(x) = alpha P_A(x) + (1-alpha) P_B(x).$$
answered Aug 28 at 14:29
HelloGoodbye
22827
2
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
add a comment |Â
up vote
1
down vote
If the PDFs of $A$, $B$ and $C$ are $P_A(x)$, $P_B(x)$ and $P_C(x)$ respectivelly, and $alpha$ is the chance that $C$ is defined as $A$ (and $1-alpha$ is the chance that $C$ is defined as $B$), then
$$P_C(x) = alpha P_A(x) + (1-alpha) P_B(x).$$
answered Aug 28 at 14:29
HelloGoodbye
22827
2
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
add a comment |Â
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If the PDFs of $A$, $B$ and $C$ are $P_A(x)$, $P_B(x)$ and $P_C(x)$ respectivelly, and $alpha$ is the chance that $C$ is defined as $A$ (and $1-alpha$ is the chance that $C$ is defined as $B$), then
$$P_C(x) = alpha P_A(x) + (1-alpha) P_B(x).$$
answered Aug 28 at 14:29
HelloGoodbye
22827
If the PDFs of $A$, $B$ and $C$ are $P_A(x)$, $P_B(x)$ and $P_C(x)$ respectivelly, and $alpha$ is the chance that $C$ is defined as $A$ (and $1-alpha$ is the chance that $C$ is defined as $B$), then
$$P_C(x) = alpha P_A(x) + (1-alpha) P_B(x).$$
answered Aug 28 at 14:29
HelloGoodbye
22827
answered Aug 28 at 14:29
HelloGoodbye
22827
answered Aug 28 at 14:29
HelloGoodbye
22827
answered Aug 28 at 14:29
HelloGoodbye
22827
22827
2
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
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This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
2
2
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
This is a good point, but don't you think it would help more to explain why this result holds, instead of just asserting it? Could you offer a simple or clear or intuitive explanation?
– whuber♦
Aug 28 at 14:33
add a comment |Â
up vote
1
down vote
This is the kind of problem where it is very helpful to use the concept of the CDF, the cumulative probability distribution function, of random variables, that totally unnecessary concept that professors drag in just to confuse students who are happy to just use pdfs.
By definition, the value of the CDF $F_X(alpha)$ of a random variable $X$ equals the probability that $X$ is no larger than the real number $alpha$, that is,
$$F_X(alpha) = PX leq alpha, ~-infty < alpha < infty.$$
Now, the law of total probability tells us that if $X$ is equally likely to be the same as a random variable $A$ or a random variable $B$, then
$$PX leq alpha = frac 12 PA leq alpha + frac 12 PB leq alpha,$$
or, in other words,
$$F_X(alpha} = frac 12 F_A(alpha} + frac 12 F_B(alpha}.$$
Remembering how your professor boringly nattered on and on about how for continuous random variables the pdf is the derivative of the CDF, we get that
$$f_X(alpha} = frac 12 f_A(alpha} + frac 12 f_B(alpha} tag1$$
which answers one of your questions. For the special case of normal random variables $A$ and $B$, can you figure out whether $(1)$ gives a normal density for $X$ or not? If you are familiar with notions such as
$$E[X] = int_-infty^infty alpha f_X(alpha} , mathrm dalpha,
tag2$$
can you figure out, by substituting the right side of $(1)$ for the $f_X(alpha)$ in $(2)$ and thinking about the expression, what $E[X]$ is in terms of $E[A]$ and $E[B]$?
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
add a comment |Â
up vote
1
down vote
This is the kind of problem where it is very helpful to use the concept of the CDF, the cumulative probability distribution function, of random variables, that totally unnecessary concept that professors drag in just to confuse students who are happy to just use pdfs.
By definition, the value of the CDF $F_X(alpha)$ of a random variable $X$ equals the probability that $X$ is no larger than the real number $alpha$, that is,
$$F_X(alpha) = PX leq alpha, ~-infty < alpha < infty.$$
Now, the law of total probability tells us that if $X$ is equally likely to be the same as a random variable $A$ or a random variable $B$, then
$$PX leq alpha = frac 12 PA leq alpha + frac 12 PB leq alpha,$$
or, in other words,
$$F_X(alpha} = frac 12 F_A(alpha} + frac 12 F_B(alpha}.$$
Remembering how your professor boringly nattered on and on about how for continuous random variables the pdf is the derivative of the CDF, we get that
$$f_X(alpha} = frac 12 f_A(alpha} + frac 12 f_B(alpha} tag1$$
which answers one of your questions. For the special case of normal random variables $A$ and $B$, can you figure out whether $(1)$ gives a normal density for $X$ or not? If you are familiar with notions such as
$$E[X] = int_-infty^infty alpha f_X(alpha} , mathrm dalpha,
tag2$$
can you figure out, by substituting the right side of $(1)$ for the $f_X(alpha)$ in $(2)$ and thinking about the expression, what $E[X]$ is in terms of $E[A]$ and $E[B]$?
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is the kind of problem where it is very helpful to use the concept of the CDF, the cumulative probability distribution function, of random variables, that totally unnecessary concept that professors drag in just to confuse students who are happy to just use pdfs.
By definition, the value of the CDF $F_X(alpha)$ of a random variable $X$ equals the probability that $X$ is no larger than the real number $alpha$, that is,
$$F_X(alpha) = PX leq alpha, ~-infty < alpha < infty.$$
Now, the law of total probability tells us that if $X$ is equally likely to be the same as a random variable $A$ or a random variable $B$, then
$$PX leq alpha = frac 12 PA leq alpha + frac 12 PB leq alpha,$$
or, in other words,
$$F_X(alpha} = frac 12 F_A(alpha} + frac 12 F_B(alpha}.$$
Remembering how your professor boringly nattered on and on about how for continuous random variables the pdf is the derivative of the CDF, we get that
$$f_X(alpha} = frac 12 f_A(alpha} + frac 12 f_B(alpha} tag1$$
which answers one of your questions. For the special case of normal random variables $A$ and $B$, can you figure out whether $(1)$ gives a normal density for $X$ or not? If you are familiar with notions such as
$$E[X] = int_-infty^infty alpha f_X(alpha} , mathrm dalpha,
tag2$$
can you figure out, by substituting the right side of $(1)$ for the $f_X(alpha)$ in $(2)$ and thinking about the expression, what $E[X]$ is in terms of $E[A]$ and $E[B]$?
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
This is the kind of problem where it is very helpful to use the concept of the CDF, the cumulative probability distribution function, of random variables, that totally unnecessary concept that professors drag in just to confuse students who are happy to just use pdfs.
By definition, the value of the CDF $F_X(alpha)$ of a random variable $X$ equals the probability that $X$ is no larger than the real number $alpha$, that is,
$$F_X(alpha) = PX leq alpha, ~-infty < alpha < infty.$$
Now, the law of total probability tells us that if $X$ is equally likely to be the same as a random variable $A$ or a random variable $B$, then
$$PX leq alpha = frac 12 PA leq alpha + frac 12 PB leq alpha,$$
or, in other words,
$$F_X(alpha} = frac 12 F_A(alpha} + frac 12 F_B(alpha}.$$
Remembering how your professor boringly nattered on and on about how for continuous random variables the pdf is the derivative of the CDF, we get that
$$f_X(alpha} = frac 12 f_A(alpha} + frac 12 f_B(alpha} tag1$$
which answers one of your questions. For the special case of normal random variables $A$ and $B$, can you figure out whether $(1)$ gives a normal density for $X$ or not? If you are familiar with notions such as
$$E[X] = int_-infty^infty alpha f_X(alpha} , mathrm dalpha,
tag2$$
can you figure out, by substituting the right side of $(1)$ for the $f_X(alpha)$ in $(2)$ and thinking about the expression, what $E[X]$ is in terms of $E[A]$ and $E[B]$?
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
answered Aug 28 at 23:21
Dilip Sarwate
28.6k248138
28.6k248138
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2
Perhaps see Wikipedia on 'mixture distribution'.
– BruceET
Aug 27 at 19:21
6
A plot could give a good hint as to whether $C$ is normally distributed.
– Kodiologist
Aug 27 at 19:24
4
Plotting the PDF of a few cases quickly shows $C$ usually is not Normal: it can have two modes. The fun part consists in obtaining a complete characterization of when $C$ is Normally distributed.
– whuber♦
Aug 27 at 20:02
3
I always find it easier to work with the CDF of a random variable than the PDF.
– BallpointBen
Aug 27 at 20:49
5
And as a hint, consider drawing someone at random from the population consisting of of all babies under one year old and all NBA players. Would you expect to find anyone who's roughly four feet tall?
– BallpointBen
Aug 27 at 20:51