$$E(e^X+Y ) = E(e^X)\times E(e^Y)$$and $$E(X^2\times Y^2) = E(X^2)\times E(Y^2),$$
where $E(\cdot)$ = expectation?
Yes, since $E(PQ)=E(P)E(Q)$ when the random variables $P$ and $Q$ space independent. In each case you can simply define brand-new random variables the are functions of the first.
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$P=e^X, Q=e^Y \implies PQ=e^X e^Y= e^X+Y$
If $X$ and also $Y$ space independent , exactly how do girlfriend knew the $X^2$ and also $Y^2$ are independent?
This deserve to be looked in ~ in two ways.
First of all, and this is what i was relying top top above, you deserve to appeal come our day-to-day understanding of exactly how the civilization works. E.g. Permit $X$ and $Y$ be the outcomes of rolling two dice. $X$ and $Y$ room independent. Now, let"s say we square every result. We clearly haven"t introduced any type of dependency by doing this, for this reason $P=X^2$ and also $Q=Y^2$ are independent.
Secondly, you have the right to look an ext deeply at the underlying stillproud.orgematics. The general an outcome is JohnK"s answer and a details instance of that is justification in korrok"s answer.
Expectation of two random variables $X$, $Y$ is characterized as the amount of the products of the worths of those random variables times your joint probabilities. For constant random variables this is
$$\stillproud.orgrmE(XY)=\int\int xy \; f_XY(x,y) \;\stillproud.orgrmdx\stillproud.orgrmdy $$
where the integrals space over the variety that $X$ and also $Y$ have the right to take, and also $f_XY$ is the share probability thickness of $X$ and also $Y$.
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In the more general situation of $\alpha(x)$ together some function of $x$ and $\beta(y)$ as some role of $y$ (you had $\alpha = \beta = x\mapsto x^2$), the expectation of your product is defined similarly.
$$\stillproud.orgrmE(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_XY(x,y) \;\stillproud.orgrmdx\stillproud.orgrmdy $$
Because $X$ and $Y$ space independent, you have the right to factorize the probability thickness $f_XY$ into the product the the probability thickness $f_X(x)$ for $X$ and the probability density $f_Y(y)$ for $Y$, i.e.: $f_XY = f_X(x)f_Y(y)$. So
$$\stillproud.orgrmE(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_X(x)f_Y(y)\;\stillproud.orgrmdx\stillproud.orgrmdy $$
Rearranging the integrand we view that the integrand is the product of terms that only depend top top $x$ and also terms that just depend on $y$ for this reason the integral itself can be split into two. Every of those 2 integrals is the an interpretation of an expectation.$$\beginalign\stillproud.orgrmE(\alpha(X)\beta(Y))&= \int\int \alpha(x)f_X(x) \; \beta(y)f_Y(y)\;\stillproud.orgrmdx\stillproud.orgrmdy \\&= \int \alpha(x)f_X(x) \;\stillproud.orgrmdx \int\beta(y)f_Y(y)\;\stillproud.orgrmdy \\&= \stillproud.orgrmE( \alpha(X))\stillproud.orgrmE(\beta(Y))\endalign$$