Calculating expectation using Gamma and Exp R.V.'s

Question

I'm working on a homework problem from Hogg (7.3.4) and am a bit stuck. I need to calculate the following expectation: $$ E(Y) = \int^{\infty}_0 \frac{2}{\theta}y e^{-\frac{y}{\theta}}(1-e^{-\frac{y}{\theta}}) dy$$

Breaking this up into 2 integrals, I get $$\int^{\infty}_0 \frac{2y}{\theta}e^{-\frac{y}{\theta}}dy - \int^{\infty}_0 \frac{2y}{\theta}e^{-\frac{2y}{\theta}}dy$$ At the point, I identify those as two Gamma distributions, so I get: $$ 2\theta E[Gamma(2, \frac{1}{\theta})] - \frac{\theta}{2}E[Gamma(2,\frac{2}{\theta})] = \frac{7\theta^2}{2} $$

However, the solution identifies the two integrals as Exponential distributions instead, and gives the solution $$ \frac{3\theta}{2} $$

But I can't see how those are exponential distributions given the y that's in front of the e. Any help would be appreciated.

Edit: I see now that $\int^{\infty}_0 \frac{2y}{\theta}e^{-\frac{y}{\theta}}dy$ after being rewritten as $2\int^{\infty}_0 \frac{y}{\theta}e^{-\frac{y}{\theta}}dy$ is 2 times the expected value of an exponential with the parameter $\frac{1}{\theta}$ which evaluates to $2\theta$. The second integral is the expected value of an exponential with parameter $\frac{2}{\theta}$ which evaluates to $\frac{\theta}{2}$. Subtract the second from the first and you get $\frac{3\theta}{2}$.

Answer 1

$3 \theta/2$ is correct. The expected value of a random variable having the Gamma distribution with parameters $2$ and $1/\theta$ would be $\int_0^\infty \dfrac{y^2}{\theta^2} e^{-y/\theta}\ dy$: one $y$ from the PDF and the other from the fact that $E[Y] = \int y f(y)\ dy$.

Answer 2

I think one way to do this is to notice that $$\int^{\infty}_0 \frac{2}{\theta}y e^{-\frac{y}{\theta}}(1-e^{-\frac{y}{\theta}}) dy=\theta\int^{\infty}_{0}2ze^{-z}(1-e^{-z})dz$$ and you can evaluate the inner integral relatively easily to be $\frac{3}{2}$. The equation holds by transforming $z=\frac{y}{\theta}$. I am at a loss what distribution this is, though. I have never seen it before.