If $X$ has an exponential distribution with mean $\theta$, does $X^2$ have mean $\theta^2$?
If not, how would I find the variance of $X^2$?
I tried this:
$$V(X^2) = E[X^4] - E[X^2]^2$$
But I'm not sure how to factor the $\theta$ into the equation. The only way I can think is if $X^2$ has expected value of $\theta^2$.
I also tried:
$$V(X^2) = 4\theta^2V(X)$$
Any help is welcome, thanks.
You can directly calculate the mean of $X^2$,suppose your exponential distribution is $f(x;\lambda)=\lambda e^{-\lambda x}$.
$E(X^2)=\int_{0}^{\infty}x^2\lambda e^{-\lambda x}dx$ by integral by parts you can get $E(X^2)=\frac{2}{\lambda^2}=2\theta^2$.
By the same method you can calculate $E(X^4)$.
You also can calculate $E(X^2)$ and $E(X^4)$ by using moment generating function if you remember:
$E(X^m)=M^{(m)}(0)$, $M$ is the mgf and $(m)$ is the order of derivative.
The mgf of the exponential distribution is $\frac{\lambda}{\lambda-t}$
The first derivative (m=1) of the mgf in term of $t$ is $\frac{\lambda}{(\lambda-t)^2}$, set $t=0$, $E(X)=\frac{1}{\lambda}=\theta$
The second derivative (m=2) of the mgf in term of $t$ is $\frac{2\lambda}{(\lambda-t)^3}$, set $t=0$, $E(X^2)=\frac{2}{\lambda^2}=2\theta^2$.
You can continue to calculate others...
Re your title -- the mean of $X^2$ and the mean of $X$ aren't even in the same units. Presumably you mean to ask something more like how the mean of $X^2$ compared to the square of the mean of $X$.
If $X$ has an exponential distribution with mean $θ$, does $X^2$ have mean $θ^2$?
(This question would be consistent with the above suggestion about a correction to the title)
The answer to this is obvious from the fact that the variance isn't zero: Given $E(X)=\theta$, if $E(X^2)=\theta^2$ then $\text{Var}(X) = E(X^2)-E(X)^2 = ...$
More generally, when $f$ is a convex function, then $E(f(X))\geq f(E(X))$ (see Jensen's inequality); equality only occurs when $f$ is linear or $\text{Var}(X)=0$.
For an exponential you can compute $E(X^k)$ quite easily, for example from the definition of expectation for a continuous pmf (the integral can readily be computed for integer $k$ - e.g. by integration by parts), though there are other approaches.
