If $X$ and $Y$ are independent random variables, are the statements below true

$$E(e^{X+Y} ) = E(e^X)\times E(e^Y)$$ and $$E(X^2\times Y^2) = E(X^2)\times E(Y^2),$$

where $E(\cdot)$ = expectation?

  • $\begingroup$ How is independency defined for you? $\endgroup$ – Stefan Hansen Feb 8 '14 at 14:29

Yes, because $E(PQ)=E(P)E(Q)$ when the random variables $P$ and $Q$ are independent. In each case you can simply define new random variables that are functions of the first.

  1. $P=e^X, Q=e^Y \implies PQ=e^X e^Y= e^{X+Y}$

  2. $P=X^2, Q=Y^2$

If $X$ and $Y$ are independent , how do you knew that $X^2$ and $Y^2$ are independent?

This can be looked at in two ways.

First of all, and this is what I was relying on above, you can appeal to our everyday understanding of how the world works. E.g. let $X$ and $Y$ be the results of rolling two dice. $X$ and $Y$ are independent. Now, let's say we square each result. We clearly haven't introduced any dependency by doing this, so $P=X^2$ and $Q=Y^2$ are independent.

Secondly, you can look more deeply at the underlying mathematics. The general result is JohnK's answer and a specific instance of that is justified in korrok's answer.

Expectation of two random variables $X$, $Y$ is defined as the sum of the products of the values of those random variables times their joint probabilities. For continuous random variables this is

$$\mathrm{E}(XY)=\int\int xy \; f_{XY}(x,y) \;\mathrm{d}x\mathrm{d}y $$

where the integrals are over the range that $X$ and $Y$ can take, and $f_{XY}$ is the joint probability density of $X$ and $Y$.

In the more general case of $\alpha(x)$ as some function of $x$ and $\beta(y)$ as some function of $y$ (you had $\alpha = \beta = x\mapsto x^2$), the expectation of their product is defined similarly.

$$\mathrm{E}(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_{XY}(x,y) \;\mathrm{d}x\mathrm{d}y $$

Because $X$ and $Y$ are independent, you can factorize the probability density $f_{XY}$ into the product of the probability density $f_X(x)$ for $X$ and the probability density $f_Y(y)$ for $Y$, i.e.: $f_{XY} = f_X(x)f_Y(y)$. So

$$\mathrm{E}(\alpha(X)\beta(Y))=\int\int \alpha(x)\beta(y) \; f_X(x)f_Y(y)\;\mathrm{d}x\mathrm{d}y $$

Rearranging the integrand we see that the integrand is the product of terms that only depend on $x$ and terms that only depend on $y$ so the integral itself can be split into two. Each of those two integrals is the definition of an expectation. $$\begin{align} \mathrm{E}(\alpha(X)\beta(Y)) &= \int\int \alpha(x)f_X(x) \; \beta(y)f_Y(y)\;\mathrm{d}x\mathrm{d}y \\ &= \int \alpha(x)f_X(x) \;\mathrm{d}x \int\beta(y)f_Y(y)\;\mathrm{d}y \\ &= \mathrm{E}( \alpha(X))\mathrm{E}(\beta(Y)) \end{align}$$

  • $\begingroup$ Could you please point a resoure to find more details about this step: "... the integrand is the product of terms that only depend on $x$ and terms that only depend on $y$ so the integral itself can be split into two." Thanks in advance. $\endgroup$ – Marcos Modenesi Jan 28 '17 at 22:46
  • 1
    $\begingroup$ @MarcosModenesi see, e.g., math.stackexchange.com/questions/1221429/…. $\endgroup$ – TooTone Jan 29 '17 at 11:13
  • $\begingroup$ Great answer!!! $\endgroup$ – information_interchange Sep 21 '18 at 20:45

Yes, both statements are correct.

In general if $X$ and $Y$ are independent:

$$E \left[ u(X)v(Y) \right]=E \left[u(X)\right]\times E\left[ v(Y)\right]$$

provided the expectations exist.


I'll just consider the case of continuous random variables with densities, which should give you enough to go on for the general case. If $X$ and $Y$ are independent random variables with densities $f_1$ and $f_2$, then their joint density can be written as a product:

$f_{X,Y}(x,y) = f_1(x)\cdot f_2(y)$

In that case,

$E[X^2\cdot Y^2] = \int\int x^2y^2f_{X,Y}(x,y)\hspace{1pt}dx\hspace{1pt}dy = \int x^2y^2f_1(x)f_2(y)\hspace{1pt}dx\hspace{1pt}dy$.

Do you see where to go from here?


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