Mixing
Finding the variance of $0.2(0.8)^x$, where $x = 0, 1, 2, 3 …$
the question asks for the expected value as well as variance of the above question.
I've attached a picture of my answer for the mean, but i can't use similar reasoning for the variance.
probability discrete-mathematics probability-distributions variance expected-value
edited Nov 21 '18 at 9:48
idea
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
add a comment |
the question asks for the expected value as well as variance of the above question.
I've attached a picture of my answer for the mean, but i can't use similar reasoning for the variance.
probability discrete-mathematics probability-distributions variance expected-value
edited Nov 21 '18 at 9:48
idea
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10
add a comment |
the question asks for the expected value as well as variance of the above question.
I've attached a picture of my answer for the mean, but i can't use similar reasoning for the variance.
probability discrete-mathematics probability-distributions variance expected-value
edited Nov 21 '18 at 9:48
idea
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
the question asks for the expected value as well as variance of the above question.
I've attached a picture of my answer for the mean, but i can't use similar reasoning for the variance.
probability discrete-mathematics probability-distributions variance expected-value
probability discrete-mathematics probability-distributions variance expected-value
edited Nov 21 '18 at 9:48
idea
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
edited Nov 21 '18 at 9:48
idea
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
edited Nov 21 '18 at 9:48
idea
2,15441025
edited Nov 21 '18 at 9:48
idea
2,15441025
edited Nov 21 '18 at 9:48
idea
2,15441025
2,15441025
asked Nov 20 '18 at 15:56
Vanessa
656
asked Nov 20 '18 at 15:56
Vanessa
656
asked Nov 20 '18 at 15:56
Vanessa
656
656
What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10
add a comment |
What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10
What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10
What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10
add a comment |
1 Answer
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$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.
$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$
$ = sum_{n=1}^infty n(1-p)^{n-1}p$
Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.
$S = 1 + 2q + 3 q^2 + ;...$
$qS = ;;;;;q + 2q^2+ ;...$
$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$
implying $S = frac{1}{(1-q)^2}$.
Using this result, we get:
$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.
$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $
$= sum_{n=1}^infty n^2(1-p)^{n-1}p $
Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.
$S = 1 + 4q + 9q^2 + ; ... $
$qS = ;;;;; q+4q^2+;... $
$(1-q)S = 1 + 3q + 5q^2 + ;...$
$q(1-q)S = ;;;; q +3q^2 + ;...$
$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$
$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $
implying $S = frac{1+q}{(1-q)^3}$.
Using this result, we get:
$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.
Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.
answered Nov 22 '18 at 8:42
Aditya Dua
80418
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.
$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$
$ = sum_{n=1}^infty n(1-p)^{n-1}p$
Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.
$S = 1 + 2q + 3 q^2 + ;...$
$qS = ;;;;;q + 2q^2+ ;...$
$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$
implying $S = frac{1}{(1-q)^2}$.
Using this result, we get:
$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.
$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $
$= sum_{n=1}^infty n^2(1-p)^{n-1}p $
Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.
$S = 1 + 4q + 9q^2 + ; ... $
$qS = ;;;;; q+4q^2+;... $
$(1-q)S = 1 + 3q + 5q^2 + ;...$
$q(1-q)S = ;;;; q +3q^2 + ;...$
$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$
$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $
implying $S = frac{1+q}{(1-q)^3}$.
Using this result, we get:
$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.
Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.
answered Nov 22 '18 at 8:42
Aditya Dua
80418
add a comment |
$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.
$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$
$ = sum_{n=1}^infty n(1-p)^{n-1}p$
Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.
$S = 1 + 2q + 3 q^2 + ;...$
$qS = ;;;;;q + 2q^2+ ;...$
$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$
implying $S = frac{1}{(1-q)^2}$.
Using this result, we get:
$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.
$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $
$= sum_{n=1}^infty n^2(1-p)^{n-1}p $
Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.
$S = 1 + 4q + 9q^2 + ; ... $
$qS = ;;;;; q+4q^2+;... $
$(1-q)S = 1 + 3q + 5q^2 + ;...$
$q(1-q)S = ;;;; q +3q^2 + ;...$
$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$
$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $
implying $S = frac{1+q}{(1-q)^3}$.
Using this result, we get:
$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.
Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.
answered Nov 22 '18 at 8:42
Aditya Dua
80418
add a comment |
$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.
$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$
$ = sum_{n=1}^infty n(1-p)^{n-1}p$
Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.
$S = 1 + 2q + 3 q^2 + ;...$
$qS = ;;;;;q + 2q^2+ ;...$
$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$
implying $S = frac{1}{(1-q)^2}$.
Using this result, we get:
$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.
$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $
$= sum_{n=1}^infty n^2(1-p)^{n-1}p $
Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.
$S = 1 + 4q + 9q^2 + ; ... $
$qS = ;;;;; q+4q^2+;... $
$(1-q)S = 1 + 3q + 5q^2 + ;...$
$q(1-q)S = ;;;; q +3q^2 + ;...$
$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$
$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $
implying $S = frac{1+q}{(1-q)^3}$.
Using this result, we get:
$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.
Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.
answered Nov 22 '18 at 8:42
Aditya Dua
80418
$P(X=n) = (1-p)^{n-1}p$, if $X$ is a geometric random variable.
$mathbb{E}[X] = sum_{n=1}^infty nP(X=n)$
$ = sum_{n=1}^infty n(1-p)^{n-1}p$
Let $S = sum_{n=1}^infty nq^{n-1}$ for some $q in (0,1)$.
$S = 1 + 2q + 3 q^2 + ;...$
$qS = ;;;;;q + 2q^2+ ;...$
$(1-q)S = 1 + q + q^2 + ;... = frac{1}{1-q}$
implying $S = frac{1}{(1-q)^2}$.
Using this result, we get:
$mathbb{E}[X] = frac{p}{(1-(1-p))^2} = frac{1}{p}$.
$mathbb{E}[X^2] = sum_{n=1}^infty n^2P(X=n) $
$= sum_{n=1}^infty n^2(1-p)^{n-1}p $
Let $S = sum_{n=1}^infty n^2 q^{n-1}$ for some $q in (0,1)$.
$S = 1 + 4q + 9q^2 + ; ... $
$qS = ;;;;; q+4q^2+;... $
$(1-q)S = 1 + 3q + 5q^2 + ;...$
$q(1-q)S = ;;;; q +3q^2 + ;...$
$(1-q)^2S = 1 + 2q + 2q^2 + ; ... = 1 + 2q(1+q+q^2+ ; ...)$
$(1-q)^2S = 1 + frac{2q}{1-q} = frac{1+q}{1-q} $
implying $S = frac{1+q}{(1-q)^3}$.
Using this result, we get:
$mathbb{E}[X^2] = frac{p(1+1-p)}{(1-(1-p))^3} = frac{2-p}{p^2} $.
Finally, $Var(X) = mathbb{E}[X^2] - mathbb{E}[X]^2 = frac{2-p}{p^2} - frac{1}{p^2} = frac{1-p}{p^2}$.
answered Nov 22 '18 at 8:42
Aditya Dua
80418
answered Nov 22 '18 at 8:42
Aditya Dua
80418
answered Nov 22 '18 at 8:42
Aditya Dua
80418
answered Nov 22 '18 at 8:42
Aditya Dua
80418
80418
add a comment |
add a comment |
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What rules do you use? If you need to find variance for F(X) rather than just the special case F(X)=X, this may help: stats.stackexchange.com/questions/5782/…
– NoChance
Nov 21 '18 at 10:10