# Averages of mixed random variables

X and Y are two random variables. X is a gaussian random variable with mean 0 and variance 4. Y is a bernoulli random variable with parameter p. Let random variables Z, H, R and L be defined as...

$$Z = X + Y$$

$$H = XY$$

$$R = X^Y$$

$$L = X^3 + Y^3$$

Determine E[Z], E[H], E[R], and E[L]

I've been stumped on this one for a while and just really don't know where to get started. I know E[X] is 0 and E[Y] should be p, but I'm not sure how to apply that to these four variables Z H R and L. Any help would definitely be appreciated!

EDIT: I believe I am allowed to assume X and Y are independent from each other.

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are $X$ and $Y$ perhaps independent? There should be some info about the relationship of one to the other... – gt6989b Oct 22 '12 at 16:42
There is nothing stated on the problem about independence but I believe unless it is explicitly stated we are allowed to assume they are independent. – Hoser Oct 22 '12 at 16:44

## 3 Answers

$E[X+Y] = E[X] + E[Y]$ and $E[X^3 + Y^3] = E[X^3] + E[Y^3]$ and you can use the individual distributions to figure out the moments.

Assuming independence,

$E[X^Y] = E[pX^1 + (1-p)X^0] = pE[X] + (1-p)$

and

$E[XY] = E[X*1*p + Y*0*q] = (1-p)E[X]$.

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$E[X+Y] = E[X]+E[Y]$ and $E[X^3+Y^3] = E[X^3] + E[Y^3]$ do not require independence. – Robert Israel Oct 22 '12 at 16:51
@RobertIsrael I agree completely. I just placed it in the wrong place, it was meant to be used for all statements below. Will edit to make clearer. – gt6989b Oct 22 '12 at 16:57
Thanks a ton, really helped me – Hoser Oct 22 '12 at 17:06

Some hints: Expected value of sum = sum of expected values.

Expected value of product of independent random variables = product of expected values.

$X^Y = X$ when $Y=1$ and $1$ when $Y=0$.

$Y^3 = Y$, while distribution of $X$ is symmetric about $0$.

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Thanks to all of you, really helped me understand this – Hoser Oct 22 '12 at 17:02

Since $X$ and $Y$ are independent we immediately get that $$EZ=EX+EY=p$$ $$EH=EX\cdot EY=0.$$ Then by properties of conditional mean we have, that $$ER=EX^Y=E_Y(E_X(X^Y|Y))=E_Y(E_X(X^0|Y=0)+E_X(X^1|Y=1))$$ then $$E_X(X^0|Y=0)=E_X(1|Y=0)=1$$ and $$E_X(X^1|Y=1)=E_X(X)=0$$ all that is left is to calculate $E_Y$, with $P(Y=0)=1-p$ and $P(Y=1)=p$ $$ER=(1-p)\cdot 1 + p\cdot 0=(1-p).$$ For $L$ i will just give you a hint $EX^p=\int_{-\infty}^{\infty}x^pf(x)dx$. Or you can use moment generating function for standard normal variable. For discrete variable, in you case Bernoulli: $$EY^3=(1-p)\cdot 0^3 + p \cdot 1^3=1.$$

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Thank you very much, helped me with understanding this – Hoser Oct 22 '12 at 17:05
The rightmost part of the third displayed equation (the line starting with $E(R)=\ldots$) is odd. – Did Oct 29 '12 at 9:26