Find distribution by using characteristic functions I can't seem to solve the following problem:
Let $I$ be a r.v. with Bernoulli distribution with parameter $p = 0.9$ and the r.v. $X$ has one point or Dirichlet distribution at zero, $\delta_{0}(dx)$ and Y has a continuous distribution s.t. $Y \sim Exponential(2)$. Given a random variable $L := IX + (1 - I)Y$, assume that $I$ is independent of $X$ and $Y$. Suppose that $L_{1}$ and $L_{2}$ are i.i.d. copies of $L$. Try to identify the distribution of $L_{1} + L_{2}$ by using characteristic functions. 
 A: Suppose that $I$ and $Y$ are independent. Without this condition the problem cannot be solved. 
Note that you need to find expected value of 
$$
e^{itL}=e^{it\,(IX+(1-I)Y)} = \begin{cases} e^{itY}, & I=0 \cr e^{itX}, & I=1 \end{cases} = (1-I)e^{itY}+Ie^{itX}
$$
So you have mixed distribution and its expectation you already know:
$$
\phi_L(t) = \mathbb E(e^{itL}) = (1-p)\mathbb E(e^{itY})+p\mathbb E(e^{itX})=(1-p)\phi_Y(t)+p\phi_X(t).
$$
Since $X=0$ a.s., $\phi_X(t)=\mathbb E(e^{itX})=1$. For exponential r.v. $Y$ with pdf 
$f_X(y)=\alpha e^{-\alpha y}\mathbb 1_{\{y\geq 0\}}$ its c.f. is 
$$
\phi_Y(t)=\frac{\alpha}{\alpha-it}=\left(1-\frac{it}{\alpha}\right)^{-1}.
$$
The c.f. of a sum of two independent r.v.'s is a product of its c.f.'s. So the c.f. of $L_1+L_2$ is 
$$
\phi_{L_1+L_2}(t) =\left((1-p)\phi_Y(t)+p\phi_X(t)\right)^2 = \left(p+(1-p)\left(1-\frac{it}{\alpha}\right)^{-1}\right)^2
$$
$$
=p^2\cdot 1+2p(1-p)\left(1-\frac{it}{\alpha}\right)^{-1}+(1-p)^2\left(1-\frac{it}{\alpha}\right)^{-2}
$$
It is a characteristic function of the mixture of three random variables: with probability $p^2$ it is $X=0$ a.s., with probability $2p(1-p)$ it is exponentially distributed $Y_1$ and with probability $(1-p)^2$ it is a sum of two independent exponentially distributed $Y_1+Y_2\sim Gamma(2,\alpha)$.
