Summation of independent discrete random variables? We have a summation of independent discrete random variables (rvs) $Y = X_1 + X_2 + \ldots + X_n$.  Assume the rvs can take non-negative real values.  How can we find the probability mass function of $Y$?
Is there any efficient method like the convolution for integer case?
 A: Use MGF:
$$
tY=t X_1 +t X_2 +\ldots t + X_n\\
e^{tY}=e^{t(X_1 +X_2 +\ldots + X_n)}\\
\varphi_Y(t)=\varphi_{X_1}(t) \cdot \varphi_{X_2}(t) \cdot \ldots \cdot \varphi_{X_n}(t)
$$
if, in addition, your $X_k$ are identically distributed, you get 
$$
\varphi_Y(t)=\varphi^{n}_{X_1}(t)
$$
A: Since the random variables are continuous, you would speak of their probability density function (instead of the probability mass function). The probability density function (PDF) of $Y$ is simply the (continuous) convolution of the PDFs of the random variables $X_i$. Convolution of two continuous random variables is defined by
$$(p_{X_1}*p_{X_2})(x)=\int_{-\infty}^{\infty}p_{X_1}(x-y)p_{X_2}(y)\;dy$$
EDIT: I was assuming your RVs are continuous, but maybe I misunderstood the question. Anyway, if they are discrete then (discrete) convolution is also the correct answer.
EXAMPLE: Let $X_1$ and $X_2$ be two discrete random variables, where $X_1$ takes on values $1/2$ and $3/4$ with probabilities $0.5$, and $X_2$ takes on values $1/8$ and $1/4$ with probabilities $0.4$ and $0.6$, respectively. So we have $p_{X_1}(x)=0.5\delta(x-1/2) + 0.5\delta(x-3/4)$ and $p_{X_2}=0.4\delta(x-1/8)+0.6\delta(x-1/4)$. Let $Y=X_1+X_2$. Then $p_Y(x)$ is given by the convolution of $p_{X_1}(x)$ and $p_{X_2}(x)$:
$$p_Y(x)=0.2\delta(x-5/8)+0.3\delta(x-3/4)+0.2\delta(x-7/8)+0.3\delta(x-1)$$
