Family of Distribution Suppose $X$ and $Y$ are independent and identically distributed random variables and let $Z=X+Y$. Then the distribution of $Z$ is in the same family as that of $X$ and $Y$ if $X$ is ??
1) Normal
2) Exponential 
3) Uniform 
4) Binomial 
I think 1) and 4) are correct.
Does the other two also have this property? 
 A: Hint
Let $X$ , $Y$ are two absolutely continues random variables with joint density function $f_{X,Y}$ and $Z=X+Y$ , we have
$${{F}_{Z}}(z)=P\,(Z\le z)=P\,(X+Y\le z)=\iint\limits_{x+y\le z}{{{f}_{X,Y}}(x\,,y)\,dy\,dx}=\int_{-\infty }^{+\infty }{\,\int_{-\infty }^{z-x}{f(x\,,y)\,dy\,dx}}$$
let $y=t-x\,$, therefore
$${{F}_{Z}}(z)=\int_{-\infty }^{+\infty }{\int_{-\infty }^{z}{f(x\,,t-x)\,dt\,dx}}=\int_{-\infty }^{z}{\int_{-\infty }^{+\infty }{f(x\,,t-x)\,dx\,dt}}$$
and
$${{f}_{Z}}(z)=\frac{d{{F}_{Z}}}{dz}=\frac{d}{dz}\,\,\int_{-\infty }^{ z}{\int_{\,-\infty }^{\,+\infty }{f(x\,,t-x)\,dx\,}\,dt}=\int_{\,-\infty }^{\,+\infty }{{{f}_{X,\,\,Y}}(x\,,z-x)\,dx}$$
$X$ , $Y$ are independent, thus

$$\large{{f}_{Z}}(z)=\int_{-\infty }^{+\infty }{{{f}_{Y}}(z-x\,})\
 {{f}_{X}}(x)\,dx$$
  Example:
   $X$ , $Y$ are independent and have exponential distribution (with $\alpha$ parameter)
  $${{f}_{Z}}(z)=\int_{-\infty }^{+\infty }{\,{{f}_{\,Y}}\,(z-x)}\ {{f}_{X}}(x)\,dx=\frac{1}{{{\alpha }^{2}}}\int_{0}^{\,z}{\,{{e}^{-\frac{x}{\alpha }}}{{e}^{-\,\frac{(z-x)}{\alpha }}}}\,dx=\frac{1}{{{\alpha }^{2}}}\,z\,{{e}^{-\frac{z}{\alpha }}}$$

A: When working with sum of i.i.d random variables, it is often convenient to use characteristic functions as $\phi_{X+Y}(t) = \phi_X(t)\phi_Y(t) = \phi_X(t)^2$.
For example, if $X,Y$ are Binomial($n,p$) then
$\phi_{X+Y}(t) = \phi_X(t)^2 = \left((1-p+pe^{it})^{n}\right)^2 = (1-p+pe^{it})^{2n}$
Hence $X+Y$ is Binomial($2n,p$).
You can use the same reasoning for the normal distribution. 
For 2), you will need to show that $\phi_X(t)^2$ cannot be written as the characteristic function of an exponential random variable. Same for 3) with uniform random variable.
