Expectation of a sum of random variables

  • #1
docnet
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Homework Statement
Given ##E[A]=0## and ##E[B]=b##, can ##E[(A+B)^2]## be simplified?
Relevant Equations
##E[A]## is the expectation of ##A##.
$$\begin{align*}
E[(A+B)^2]&=E[A^2+2AB+B^2]\\
&=E[A^2]+2E[AB]+E[B^2]\\
&=2E[AB]+E[B^2].
\end{align*}$$

Can the terms ##2E[AB]## and ##E[B^2]## be simplified any more? Thanks, friends.
 
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  • #2
How did you estimate E[A^2]?
E[ AB ]=E[ A ]E[ B ]
 
  • #3
anuttarasammyak said:
E[AB]=E[A]E How did you estimate ##E[A^2]##?
Oh I have made a mistake.. we do not know that ##E[A^2]=0##.

If ##A## takes values ##-1## and ##1## with equal probability then ##A^2=1## with probability ##1##. So ##E[A^2]=1.##.

Thank you!
 
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  • #4
anuttarasammyak said:
How did you estimate E[A^2]?
E[ AB ]=E[ A ]E[ B ]

This is only valid if [itex]A[/itex] and [itex]B[/itex] are independent (it doesn't hold, for example, when [itex]A = B[/itex]), and we are not told that they are.

Using [tex]\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2[/tex], we have [tex]
\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.[/tex]
 
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  • #5
pasmith said:
This is only valid if [itex]A[/itex] and [itex]B[/itex] are independent (it doesn't hold, for example, when [itex]A = B[/itex]), and we are not told that they are.

Using [tex]\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2[/tex], we have [tex]
\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.[/tex]
That is extremely useful ! Thanks.
 
  • #6
pasmith said:
This is only valid if [itex]A[/itex] and [itex]B[/itex] are independent (it doesn't hold, for example, when [itex]A = B[/itex]), and we are not told that they are.

Using [tex]\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2[/tex], we have [tex]
\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.[/tex]
I can think further of your example when ##A=B \sim~N(0,1)##, so that ##A^2 \~ sim mathbb Chi^2(1)##, with expectation/expected value ##1##, while ##E[A]=0##, so that ##E[A^2]=1 \neq E[A]E[A]=0.0=0##
 
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  • #7
pasmith said:
This is only valid if [itex]A[/itex] and [itex]B[/itex] are independent (it doesn't hold, for example, when [itex]A = B[/itex]), and we are not told that they are.

Using [tex]\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2[/tex], we have [tex]
\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.[/tex]
I can think further of your example when ##A=B \sim\mathbb N(0,1)##, so that ##A^2 \sim \mathbb{ \chi^2(1)}##, with expectation/expected value ##1##, while ##E[A]=0##, so that ##E[A^2]=1 \neq E[A]E[A]=0.0=0##
 
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  • #8
WWGD said:
I can think further of your example when ##A=B ## ~##\mathbb N(0,1)##, so that ##A^2 ## ~##\ mathbb \chi^2(1)##, with expectation/expected value ##1##, while ##E[A]=0##, so that ##E[A^2]=1 \neq E[A]E[A]=0.0=##
Try ##\sim##
 
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  • #9
Orodruin said:
Try ##\sim##
Thanks, fixed it. Phew! Need an upgrade and refresher on my Tex.
 
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  • #10
What is it meant by simplified? $$ E [ ( A + B ) ^ 2 ] $$ is simpler than $$ E [ A ^ 2 ] + 2 E [ A B ] + E [ B ^ 2 ] $$ and demands less calculation. Only in the case where A and B are independent $$ E [ ( A + B ) ^ 2 ] $$ can become a simpler expression which is $$ E [ A ^ 2 ] + E [ B ^ 2 ] $$ and which demands less calculation.
 
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  • #11
Gavran said:
What is it meant by simplified? $$ E [ ( A + B ) ^ 2 ] $$ is simpler than $$ E [ A ^ 2 ] + 2 E [ A B ] + E [ B ^ 2 ] $$ and demands less calculation. Only in the case where A and B are independent $$ E [ ( A + B ) ^ 2 ] $$ can become a simpler expression which is $$ E [ A ^ 2 ] + E [ B ^ 2 ] $$ and which demands less calculation.

[itex]A[/itex] and [itex]B[/itex] being independent does not imply that [itex]E[AB] = 0[/itex].
 
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  • #12
pasmith said:
[itex]A[/itex] and [itex]B[/itex] being independent does not imply that [itex]E[AB] = 0[/itex].
It does if you pair it with ##E[A]=0## as given in the OP.
 
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