Help on Moment Generating Functions
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I have recently been given a set of practice problems for my probabilities course and I have no idea where to even start on this question.
The distribution of X = the number of toppings ordered by a randomly selected customer is given in the table below. It turns out that X is independent of the size of the pizza and the type of cheese and that each topping is equally popular as are the two cheese types.
P(X=x)
0 = 0.3
1 = 0.3
2 = 0.1
3 = 0.1
4 = 0.2
What is the mgf of X?
I am pretty sure this is a Binomial Distribution. So would I just put $[(1-ø) + øe^t]n$ as my answer?
Overall I am really confused on this topic and would like some help that isn't too discrete. Thank You!
mgf
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up vote
1
down vote
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I have recently been given a set of practice problems for my probabilities course and I have no idea where to even start on this question.
The distribution of X = the number of toppings ordered by a randomly selected customer is given in the table below. It turns out that X is independent of the size of the pizza and the type of cheese and that each topping is equally popular as are the two cheese types.
P(X=x)
0 = 0.3
1 = 0.3
2 = 0.1
3 = 0.1
4 = 0.2
What is the mgf of X?
I am pretty sure this is a Binomial Distribution. So would I just put $[(1-ø) + øe^t]n$ as my answer?
Overall I am really confused on this topic and would like some help that isn't too discrete. Thank You!
mgf
New contributor
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have recently been given a set of practice problems for my probabilities course and I have no idea where to even start on this question.
The distribution of X = the number of toppings ordered by a randomly selected customer is given in the table below. It turns out that X is independent of the size of the pizza and the type of cheese and that each topping is equally popular as are the two cheese types.
P(X=x)
0 = 0.3
1 = 0.3
2 = 0.1
3 = 0.1
4 = 0.2
What is the mgf of X?
I am pretty sure this is a Binomial Distribution. So would I just put $[(1-ø) + øe^t]n$ as my answer?
Overall I am really confused on this topic and would like some help that isn't too discrete. Thank You!
mgf
New contributor
I have recently been given a set of practice problems for my probabilities course and I have no idea where to even start on this question.
The distribution of X = the number of toppings ordered by a randomly selected customer is given in the table below. It turns out that X is independent of the size of the pizza and the type of cheese and that each topping is equally popular as are the two cheese types.
P(X=x)
0 = 0.3
1 = 0.3
2 = 0.1
3 = 0.1
4 = 0.2
What is the mgf of X?
I am pretty sure this is a Binomial Distribution. So would I just put $[(1-ø) + øe^t]n$ as my answer?
Overall I am really confused on this topic and would like some help that isn't too discrete. Thank You!
mgf
mgf
New contributor
New contributor
edited 1 hour ago
SecretAgentMan
293114
293114
New contributor
asked 2 hours ago
Dillon Hector
153
153
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New contributor
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add a comment |Â
1 Answer
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4
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$X$ is not a binomial distribution. A binomial distribution is the number of successes $X$ out of $n$ independent trials with constant probability of success $p$.
Now to answer your question, you're given the PMF of your random variable $X$ explicitly. The moment generating function of a random variable $X$ is defined as
$$M_X (t) = E[e^Xt]$$
And for a discrete random variable $Z$, expectation of $g(Z)$ is defined as
$E[g(Z)] = sum_z g(z) times P(Z=z)$
Therefore, the moment generating function of your random variable $X$ is defined as
$M_X(t) = sum_x e^xt times P(X=x) $
$= e^0ttimes 0.3 + e^1ttimes 0.3 + e^2ttimes 0.1 + e^3ttimes 0.1 + e^4t times 0.2$
Which is clearly a function in $t$ as you would expect
New contributor
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$X$ is not a binomial distribution. A binomial distribution is the number of successes $X$ out of $n$ independent trials with constant probability of success $p$.
Now to answer your question, you're given the PMF of your random variable $X$ explicitly. The moment generating function of a random variable $X$ is defined as
$$M_X (t) = E[e^Xt]$$
And for a discrete random variable $Z$, expectation of $g(Z)$ is defined as
$E[g(Z)] = sum_z g(z) times P(Z=z)$
Therefore, the moment generating function of your random variable $X$ is defined as
$M_X(t) = sum_x e^xt times P(X=x) $
$= e^0ttimes 0.3 + e^1ttimes 0.3 + e^2ttimes 0.1 + e^3ttimes 0.1 + e^4t times 0.2$
Which is clearly a function in $t$ as you would expect
New contributor
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
$X$ is not a binomial distribution. A binomial distribution is the number of successes $X$ out of $n$ independent trials with constant probability of success $p$.
Now to answer your question, you're given the PMF of your random variable $X$ explicitly. The moment generating function of a random variable $X$ is defined as
$$M_X (t) = E[e^Xt]$$
And for a discrete random variable $Z$, expectation of $g(Z)$ is defined as
$E[g(Z)] = sum_z g(z) times P(Z=z)$
Therefore, the moment generating function of your random variable $X$ is defined as
$M_X(t) = sum_x e^xt times P(X=x) $
$= e^0ttimes 0.3 + e^1ttimes 0.3 + e^2ttimes 0.1 + e^3ttimes 0.1 + e^4t times 0.2$
Which is clearly a function in $t$ as you would expect
New contributor
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$X$ is not a binomial distribution. A binomial distribution is the number of successes $X$ out of $n$ independent trials with constant probability of success $p$.
Now to answer your question, you're given the PMF of your random variable $X$ explicitly. The moment generating function of a random variable $X$ is defined as
$$M_X (t) = E[e^Xt]$$
And for a discrete random variable $Z$, expectation of $g(Z)$ is defined as
$E[g(Z)] = sum_z g(z) times P(Z=z)$
Therefore, the moment generating function of your random variable $X$ is defined as
$M_X(t) = sum_x e^xt times P(X=x) $
$= e^0ttimes 0.3 + e^1ttimes 0.3 + e^2ttimes 0.1 + e^3ttimes 0.1 + e^4t times 0.2$
Which is clearly a function in $t$ as you would expect
New contributor
$X$ is not a binomial distribution. A binomial distribution is the number of successes $X$ out of $n$ independent trials with constant probability of success $p$.
Now to answer your question, you're given the PMF of your random variable $X$ explicitly. The moment generating function of a random variable $X$ is defined as
$$M_X (t) = E[e^Xt]$$
And for a discrete random variable $Z$, expectation of $g(Z)$ is defined as
$E[g(Z)] = sum_z g(z) times P(Z=z)$
Therefore, the moment generating function of your random variable $X$ is defined as
$M_X(t) = sum_x e^xt times P(X=x) $
$= e^0ttimes 0.3 + e^1ttimes 0.3 + e^2ttimes 0.1 + e^3ttimes 0.1 + e^4t times 0.2$
Which is clearly a function in $t$ as you would expect
New contributor
edited 3 mins ago
New contributor
answered 1 hour ago
Xiaomi
905
905
New contributor
New contributor
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
add a comment |Â
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
So exactly what type of distribution would X be? If any.
â Dillon Hector
1 hour ago
1
1
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
It's not a specific type, it's described by the probability function given to you. So long as the function's probabilities sum to 1, it's a valid discrete distribution, it doesn't need to be of a specific type.
â Xiaomi
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Okay, so say I did get a Binomial Distribution, would I use the formula and then that is it? Or would I do what was done above? Sorry just a bit confused.
â Dillon Hector
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
Yes. If you know $X$ has a Binomial$(n,p)$ distribution, then you can use the formula for it's MGF rather than deriving it. In the case of parameters $n,p$, it would have MGF $(1-p+pe^t)^n$. You just need to make sure the binomial assumptions are actually true, or if you are explitly told in the question. No need to apologize, you're here to learn after all!
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
To further add: if you are explicitly told in the question "$X$ has ___ distribution with parameters ___" then you can just look up the distribution's wikipedia page and find it's MGF formula on the side bar. As long as you know the parameters, you can just chuck them in to the MGF formular right away
â Xiaomi
1 hour ago
add a comment |Â
Dillon Hector is a new contributor. Be nice, and check out our Code of Conduct.
Dillon Hector is a new contributor. Be nice, and check out our Code of Conduct.
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