Random independent variables, a question of expected value

Question

The density function of $X$ and $Y$ (two independent variables) are respectively : $$\phi_X(x)=\begin{cases} \frac{1}{2}(1+x) , x \in (-1,1) \\ 0, \text{otherwise} \end{cases}$$

$$\phi_Y(y)=\begin{cases} 2y , y \in (0,1) \\ 0, \text{otherwise} \end{cases}$$

$Z=XY$

Find: $EZ^{2014}$ Expected value. So I can definitely find the PDF and CDF of Z and I have the formula for expected vale of Z, or and absolutely continuous variable: $$\int_{-\infty}^{+ \infty}zdF_Z(z)$$, where $F$ is the CDF of variable $Z$, what about the $2014$ degree of the variable ?

Answer 1

Recall that $$E[g(X)]=\int_{-\infty}^{\infty}g(x)\cdot f_X(x)~dx$$

So simply your expectattion will be $$E[Z^{2014}]=\int_{-\infty}^{\infty}z^{2014}\cdot f_Z(x)~dz$$

Answer 2

I would start as follows $E[Z^{2014}] = E[X^{2014}] E[Y^{2014}]$

Answer 3

Both @A.D and @wiskundeliefhebber have the right idea here. $X$ and $Y$ are independent therefore $X^{2014}$ and $Y^{2014}$ are independent so $\mathbb E Z^{2014} \equiv \mathbb E(XY)^{2014} = \mathbb E X^{2014}\mathbb E Y^{2014}$ and you can determine these in one of two ways:

• Find the moment generating functions of $X$ and $Y$ as the expectations above are both moments
• Evaluate the expectations directly. You should find that the integrals aren't actually very hard even if they look like they might be