Expectation of a sum of random variables

[docnet]

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.

 

[anuttarasammyak]

How did you estimate E[A^2]?
E[ AB ]=E[ A ]E[ B ]

 

[docnet]

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!

 

[pasmith]

How did you estimate E[A^2]?
E[ AB ]=E[ A ]E[ B ]

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

Using $$\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$, we have $$\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.$$

 

[docnet]

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

Using $$\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$, we have $$\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.$$

That is extremely useful ! Thanks.

 

[WWGD]

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

Using $$\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$, we have $$\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.$$

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 $E[A^2]=1 \neq E[A]E[A]=0.0=0$

 

[WWGD]

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

Using $$\newcommand{\Var}{\operatorname{Var}}\Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2$$, we have $$\mathbb{E}[(A + B)^2] = \Var(A + B) + (\mathbb{E}[A + B])^2 = \Var(A + B) + b^2.$$

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 $E[A^2]=1 \neq E[A]E[A]=0.0=0$

 

[Orodruin]

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 $\sim$

 

[WWGD]

Try $\sim$

Thanks, fixed it. Phew! Need an upgrade and refresher on my Tex.

 

[Gavran]

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.

 

[pasmith]

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.

$A$ and $B$ being independent does not imply that $E[AB] = 0$.

 

[Orodruin]

$A$ and $B$ being independent does not imply that $E[AB] = 0$.

It does if you pair it with $E[A]=0$ as given in the OP.