If X is a binomial random variable, find $E(X(X - 1) (X - 2))$ [duplicate]
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Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$
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If X is a binomial random variable, how do you find $E(X(X - 1)(X - 2))$?
Here's my approach:
$E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)$
$E(X) = np, Var(X) = np(1 - p)$
$Var(X) = E(X^2) - (E(X))^2$
$E(X^2) = np(1 - p) + (np)^2$
So, $E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)$
How should I deal with $E(X^3)$?
probability probability-theory probability-distributions
edited Nov 21 at 16:15
Key Flex
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
marked as duplicate by Clement C., Math Lover, drhab probability
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Nov 21 at 16:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$
4 answers
If X is a binomial random variable, how do you find $E(X(X - 1)(X - 2))$?
Here's my approach:
$E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)$
$E(X) = np, Var(X) = np(1 - p)$
$Var(X) = E(X^2) - (E(X))^2$
$E(X^2) = np(1 - p) + (np)^2$
So, $E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)$
How should I deal with $E(X^3)$?
probability probability-theory probability-distributions
edited Nov 21 at 16:15
Key Flex
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
marked as duplicate by Clement C., Math Lover, drhab probability
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Nov 21 at 16:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21
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up vote
0
down vote
favorite
This question already has an answer here:
Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$
4 answers
If X is a binomial random variable, how do you find $E(X(X - 1)(X - 2))$?
Here's my approach:
$E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)$
$E(X) = np, Var(X) = np(1 - p)$
$Var(X) = E(X^2) - (E(X))^2$
$E(X^2) = np(1 - p) + (np)^2$
So, $E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)$
How should I deal with $E(X^3)$?
probability probability-theory probability-distributions
edited Nov 21 at 16:15
Key Flex
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
This question already has an answer here:
Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$
4 answers
If X is a binomial random variable, how do you find $E(X(X - 1)(X - 2))$?
Here's my approach:
$E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)$
$E(X) = np, Var(X) = np(1 - p)$
$Var(X) = E(X^2) - (E(X))^2$
$E(X^2) = np(1 - p) + (np)^2$
So, $E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)$
How should I deal with $E(X^3)$?
This question already has an answer here:
Given that $X sim operatorname{Binomial}(n,p)$, Find $mathbb{E}[X(X-1)(X-2)(X-3)]$
4 answers
probability probability-theory probability-distributions
probability probability-theory probability-distributions
edited Nov 21 at 16:15
Key Flex
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
edited Nov 21 at 16:15
Key Flex
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
edited Nov 21 at 16:15
Key Flex
7,01431229
edited Nov 21 at 16:15
Key Flex
7,01431229
edited Nov 21 at 16:15
Key Flex
7,01431229
7,01431229
asked Nov 21 at 16:12
Yolanda Hui
186
asked Nov 21 at 16:12
Yolanda Hui
186
asked Nov 21 at 16:12
Yolanda Hui
186
186
marked as duplicate by Clement C., Math Lover, drhab probability
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Nov 21 at 16:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Nov 21 at 16:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21
add a comment |
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21
add a comment |
1 Answer
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Given any discrete random variable $X$, the probability generating function (pgf) of $X$ is $G_X(t):=mathbb{E}t^X$. Note that $mathbb{E}prod_{i=0}^{n-1}(X-i)=G_X^{(n)}(1)$. As an example consider $Xsim B(n,,p)implies G_X(t)=sum_{k=0}^inftybinom{n}{k}(pt)^k q^{n-k}=(q+pt)^n$. Hence $mathbb{E}X(X-1)(X-2)=n(n-1)(n-2)p^3(q+pt)^{n-3}|_{t=1}=n(n-1)(n-2)p^3$.
answered Nov 21 at 16:17
J.G.
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1 Answer
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1 Answer
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active
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active
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up vote
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Given any discrete random variable $X$, the probability generating function (pgf) of $X$ is $G_X(t):=mathbb{E}t^X$. Note that $mathbb{E}prod_{i=0}^{n-1}(X-i)=G_X^{(n)}(1)$. As an example consider $Xsim B(n,,p)implies G_X(t)=sum_{k=0}^inftybinom{n}{k}(pt)^k q^{n-k}=(q+pt)^n$. Hence $mathbb{E}X(X-1)(X-2)=n(n-1)(n-2)p^3(q+pt)^{n-3}|_{t=1}=n(n-1)(n-2)p^3$.
answered Nov 21 at 16:17
J.G.
19k21932
add a comment |
up vote
0
down vote
Given any discrete random variable $X$, the probability generating function (pgf) of $X$ is $G_X(t):=mathbb{E}t^X$. Note that $mathbb{E}prod_{i=0}^{n-1}(X-i)=G_X^{(n)}(1)$. As an example consider $Xsim B(n,,p)implies G_X(t)=sum_{k=0}^inftybinom{n}{k}(pt)^k q^{n-k}=(q+pt)^n$. Hence $mathbb{E}X(X-1)(X-2)=n(n-1)(n-2)p^3(q+pt)^{n-3}|_{t=1}=n(n-1)(n-2)p^3$.
answered Nov 21 at 16:17
J.G.
19k21932
add a comment |
up vote
0
down vote
up vote
0
down vote
Given any discrete random variable $X$, the probability generating function (pgf) of $X$ is $G_X(t):=mathbb{E}t^X$. Note that $mathbb{E}prod_{i=0}^{n-1}(X-i)=G_X^{(n)}(1)$. As an example consider $Xsim B(n,,p)implies G_X(t)=sum_{k=0}^inftybinom{n}{k}(pt)^k q^{n-k}=(q+pt)^n$. Hence $mathbb{E}X(X-1)(X-2)=n(n-1)(n-2)p^3(q+pt)^{n-3}|_{t=1}=n(n-1)(n-2)p^3$.
answered Nov 21 at 16:17
J.G.
19k21932
Given any discrete random variable $X$, the probability generating function (pgf) of $X$ is $G_X(t):=mathbb{E}t^X$. Note that $mathbb{E}prod_{i=0}^{n-1}(X-i)=G_X^{(n)}(1)$. As an example consider $Xsim B(n,,p)implies G_X(t)=sum_{k=0}^inftybinom{n}{k}(pt)^k q^{n-k}=(q+pt)^n$. Hence $mathbb{E}X(X-1)(X-2)=n(n-1)(n-2)p^3(q+pt)^{n-3}|_{t=1}=n(n-1)(n-2)p^3$.
answered Nov 21 at 16:17
J.G.
19k21932
answered Nov 21 at 16:17
J.G.
19k21932
answered Nov 21 at 16:17
J.G.
19k21932
answered Nov 21 at 16:17
J.G.
19k21932
19k21932
add a comment |
add a comment |
math.stackexchange.com/questions/1476676/…
– giannispapav
Nov 21 at 16:17
Analogue/Duplicate of this question from yesterday. Is it recent homework? (and Why did you leave the hint/"following steps" suggested out, incidentally?)
– Clement C.
Nov 21 at 16:21