Questions on probability distributions (1) Let $X_1,X_2$ be two independent gamma-distributed random variables: $X_1 \sim \Gamma(r,1), X_2 \sim \Gamma(s,1)$. 
Are $Z_1:=\frac{X_1}{X_1+X_2}$ and $Z_2:= X_1+X_2$ independent? if yes, I have to find their density. I have already found that $X_1=Y_1Y_2$ and $X_2=Y_1(1-Y_2)$. But I am not done. What is the domain of $Y_1$ and $Y_2$? Since $X_1,X_2>0$ I have that $Y_1>0$ and $0<Y_2<1$.
(2) If $X_1 \sim B (a,b), X_2 \sim B(a+b,c),$ prove $X_1 X_2 \sim B(a,b+c)$
(3) If $X \sim N(0,\sigma^2),$ calculate $E(X^n).$ What I know is that
$$E(x^n) = \int_{-\infty}^{\infty}x^n\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}\;dx$$
I've tried solving numerous times it by parts and then taking limits but I keep getting $0$ and not $3t^2$! 
Can somebody give me a better direction?
 A: The integral $\displaystyle \int_{-\infty}^\infty x^n\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}\;dx$ is in fact $0$ when $n$ is odd, since it's the integral of an odd function over an interval that is symmetric about $0.$ (If the integrals of the positive and negative parts were both infinite, then complications would arise, but we don't have that problem in this case.
Here is what I suspect you did:
Let
\begin{align}
& u = \dfrac {x^2} {2t} \\[6pt]
& t\, du = x \, dx \\[6pt]
& x^n = \big( 2tu \big)^{n/2}
\end{align}
Then as $x$ goes from $-\infty$ to $+\infty,$ $u$ goes from $+\infty$ down to $0$ and back up to $+\infty,$ so you get
$$
\int_{+\infty}^{+\infty}
$$
and you conclude that that is $0.$
But you shouldn't use a non-one-to-one substitution.
Instead, write
$$
\int_{-\infty}^{+\infty} x^n \frac 1 {\sqrt{2\pi t}} e^{-x^2/(2t)} \, dx = 2\int_0^{+\infty} x^n \frac 1 {\sqrt{2\pi t}} e^{-x^2/(2t)} \, dx
$$
i.e.
$$
\int_{-\infty}^{+\infty} = 2\int_0^{+\infty}.
$$
This is correct when $n$ is even. Then go on from there, using the substitution above.
Postscript: With $n=4,$ we have
$$
x^4 \,dx = x^3\big(x\,dx\big) = (2tu)^{3/2} \big(t\,du\big)
$$
and so
\begin{align}
& 2\int_0^{+\infty} x^3 \frac 1 {\sqrt{2\pi t}} e^{-x^2/(2t)} \big(x \, dx\big) \\[8pt]
= {} & \frac 2 {\sqrt{2\pi t}} \int_0^{+\infty} (2tu)^{3/2} e^{-u} \big(t\,du\big) \\[8pt]
= {} & \frac 2 {\sqrt{2\pi t}} \cdot (2t)^{3/2} \cdot t \int_0^{+\infty} u^{3/2} e^{-u} \, du \\[8pt]
= {} & \frac 4 {\sqrt{\pi t}} \cdot t^{5/2} \Gamma\left( \frac 5 2 \right) \tag 1 \\[8pt]
= {} & \frac{4t^2}{\sqrt\pi} \cdot\frac 1 2 \cdot \frac 3 2 \Gamma\left( \frac 1 2 \right) \\[8pt]
= {} & 3t^2.
\end{align}
Starting on line $(1)$ you need to know some properties of the Gamma function.
A: 
If $X_1 \sim B (a,b), X_2 \sim B(a+b,c),$ prove $X_1 X_2 \sim B(a,b+c)$

Suppose $Y_1,Y_2,Y_3$ are independent and have gamma distributions
$$
\text{constant} \times x^{k-1} e^{-x} \, dx \quad \text{for } x \ge0
$$
for $k=a,b,c$ respectively. Then the distribution of $X_1$ is the same as that of $Y_1/(Y_1+Y_2),$ and the distribution of $X_2$ is the same as that of $(Y_1+Y_2)/(Y_1+Y_2+Y_3).$
If the joint distribution of $(X_1,X_2),$ rather than only the two marginal distributions, is the same as that of $\big( Y_1, Y_1+Y_2\big)/\big(Y_1+Y_2+Y_3\big),$ then you can see  how the result would follow. But that is a big "if" and that information is not given in your question.
