Mixing
Variance of Coin Flips Until H
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If we flip a fair coin until we get heads, what is the variance of the number of flips to do this?
My attempt is:
$$E(flips):=Y=1times P(H)+(1+Y)times P(T)$$
$$Rightarrow Y=frac12+frac1+Y2$$
$$Rightarrow Y=2$$
I know this is correct. Now we attempt to compute:
$$E(flips^2):=X=1^2times P(H)+(sqrtX+1)^2times P(T)$$
$$Rightarrow X=frac12+frac(sqrtX+1)^22$$
$$Rightarrow X=2(2+sqrt 3)$$
I know the right answer is $E(flips^2)=6$, but I'm not sure how to solve it using this recursive strategy, as the above seemed most natural to me.*
*ie take the expected value $X$, square root it to get the number of flips (not squared), add one, then square it again.
variance expected-value
asked 3 hours ago
Dan
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up vote
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favorite
If we flip a fair coin until we get heads, what is the variance of the number of flips to do this?
My attempt is:
$$E(flips):=Y=1times P(H)+(1+Y)times P(T)$$
$$Rightarrow Y=frac12+frac1+Y2$$
$$Rightarrow Y=2$$
I know this is correct. Now we attempt to compute:
$$E(flips^2):=X=1^2times P(H)+(sqrtX+1)^2times P(T)$$
$$Rightarrow X=frac12+frac(sqrtX+1)^22$$
$$Rightarrow X=2(2+sqrt 3)$$
I know the right answer is $E(flips^2)=6$, but I'm not sure how to solve it using this recursive strategy, as the above seemed most natural to me.*
*ie take the expected value $X$, square root it to get the number of flips (not squared), add one, then square it again.
variance expected-value
asked 3 hours ago
Dan
82
New contributor
Dan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we flip a fair coin until we get heads, what is the variance of the number of flips to do this?
My attempt is:
$$E(flips):=Y=1times P(H)+(1+Y)times P(T)$$
$$Rightarrow Y=frac12+frac1+Y2$$
$$Rightarrow Y=2$$
I know this is correct. Now we attempt to compute:
$$E(flips^2):=X=1^2times P(H)+(sqrtX+1)^2times P(T)$$
$$Rightarrow X=frac12+frac(sqrtX+1)^22$$
$$Rightarrow X=2(2+sqrt 3)$$
I know the right answer is $E(flips^2)=6$, but I'm not sure how to solve it using this recursive strategy, as the above seemed most natural to me.*
*ie take the expected value $X$, square root it to get the number of flips (not squared), add one, then square it again.
variance expected-value
asked 3 hours ago
Dan
82
New contributor
Dan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we flip a fair coin until we get heads, what is the variance of the number of flips to do this?
My attempt is:
$$E(flips):=Y=1times P(H)+(1+Y)times P(T)$$
$$Rightarrow Y=frac12+frac1+Y2$$
$$Rightarrow Y=2$$
I know this is correct. Now we attempt to compute:
$$E(flips^2):=X=1^2times P(H)+(sqrtX+1)^2times P(T)$$
$$Rightarrow X=frac12+frac(sqrtX+1)^22$$
$$Rightarrow X=2(2+sqrt 3)$$
I know the right answer is $E(flips^2)=6$, but I'm not sure how to solve it using this recursive strategy, as the above seemed most natural to me.*
*ie take the expected value $X$, square root it to get the number of flips (not squared), add one, then square it again.
variance expected-value
variance expected-value
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Dan
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2 Answers
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You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails. We will compute $E(N^2)$ using the Law of Total Expectation:
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$. Let's find its distribution. Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$. Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$. Thus
beginalign E(N^2) & = E(E(N^2 | H_1)) \
& = 1^2 cdot frac12 + (E(N^2) + 2E(N) + 1)cdot 1/2 \
& = frac12 + (E(N^2) + 2(2) + 1)cdot 1/2. \
endalign
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
answered 1 hour ago
Gordon Honerkamp-Smith
1546
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Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=frac1-pp^2$ and $E(flips)=frac1p$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
answered 1 hour ago
OUrista
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails. We will compute $E(N^2)$ using the Law of Total Expectation:
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$. Let's find its distribution. Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$. Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$. Thus
beginalign E(N^2) & = E(E(N^2 | H_1)) \
& = 1^2 cdot frac12 + (E(N^2) + 2E(N) + 1)cdot 1/2 \
& = frac12 + (E(N^2) + 2(2) + 1)cdot 1/2. \
endalign
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
answered 1 hour ago
Gordon Honerkamp-Smith
1546
add a comment |Â
up vote
3
down vote
accepted
You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails. We will compute $E(N^2)$ using the Law of Total Expectation:
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$. Let's find its distribution. Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$. Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$. Thus
beginalign E(N^2) & = E(E(N^2 | H_1)) \
& = 1^2 cdot frac12 + (E(N^2) + 2E(N) + 1)cdot 1/2 \
& = frac12 + (E(N^2) + 2(2) + 1)cdot 1/2. \
endalign
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
answered 1 hour ago
Gordon Honerkamp-Smith
1546
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails. We will compute $E(N^2)$ using the Law of Total Expectation:
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$. Let's find its distribution. Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$. Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$. Thus
beginalign E(N^2) & = E(E(N^2 | H_1)) \
& = 1^2 cdot frac12 + (E(N^2) + 2E(N) + 1)cdot 1/2 \
& = frac12 + (E(N^2) + 2(2) + 1)cdot 1/2. \
endalign
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
answered 1 hour ago
Gordon Honerkamp-Smith
1546
You are essentially using the Law of Total Expectation; see the accepted answer here for a review: https://math.stackexchange.com/questions/521609/finding-expected-value-with-recursion.
Let $N$ be the number of flips required until the first heads appears, and let $H_1$ be the indicator of a heads on the first flip; i.e. $H_1 = 1$ if the first flip is heads and $H_1 = 0$ if it is tails. We will compute $E(N^2)$ using the Law of Total Expectation:
$$E(N^2) = E(E(N^2 | H_1)).$$
The conditional expectation $E(N^2 | H_1)$ is a random variable; in particular it is a function of $H_1$. Let's find its distribution. Conditional on $H_1 = 1$ (i.e. when the first flip is heads), the number of flips until heads appears will of course be one, so $E(N^2 | H_1 = 1) = 1^2$. Conditional on $H_1 = 0$ (when the first flip is tails), the number of flips until heads appears will be one more than in the unconditional case, hence the conditional expectation is $E(N^2 | H_1 = 0) = E((N + 1)^2) = E(N^2) + 2E(N) + 1$. Thus
beginalign E(N^2) & = E(E(N^2 | H_1)) \
& = 1^2 cdot frac12 + (E(N^2) + 2E(N) + 1)cdot 1/2 \
& = frac12 + (E(N^2) + 2(2) + 1)cdot 1/2. \
endalign
Solving the equation, we find that $E(N^2) = 6$ and then $Var(N) = 6 - 2^2 = 2$.
answered 1 hour ago
Gordon Honerkamp-Smith
1546
edited 59 mins ago
answered 1 hour ago
Gordon Honerkamp-Smith
1546
answered 1 hour ago
Gordon Honerkamp-Smith
1546
answered 1 hour ago
Gordon Honerkamp-Smith
1546
1546
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up vote
1
down vote
Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=frac1-pp^2$ and $E(flips)=frac1p$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
answered 1 hour ago
OUrista
112
New contributor
OUrista is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=frac1-pp^2$ and $E(flips)=frac1p$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
answered 1 hour ago
OUrista
112
New contributor
OUrista is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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up vote
1
down vote
up vote
1
down vote
Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=frac1-pp^2$ and $E(flips)=frac1p$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
answered 1 hour ago
OUrista
112
New contributor
OUrista is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Although this doesn't follow your recursive approach, you can use the geometric distribution's properties here.
You have $X$, which I will refer to as $flips$ from now on, is distributed according to $Geometric(p=0.5)$.
Recall the definition of Variance, $Var(flips)=E(flips^2) + [E(flips)]^2$
where,
$Var(flips)=frac1-pp^2$ and $E(flips)=frac1p$.
Thus, $E(flips^2)=Var(flips) + [E(flips)]^2 = 2+4 = 6$.
answered 1 hour ago
OUrista
112
New contributor
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edited 56 mins ago
answered 1 hour ago
OUrista
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answered 1 hour ago
OUrista
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answered 1 hour ago
OUrista
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112
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Dan is a new contributor. Be nice, and check out our Code of Conduct.
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