So I'm preparing for a reexamination for an introductory statistics course, last time I had trouble finding a way to specify the distribution of variables/vectors. So my question is, maybe a more general way of specifying what distribution they have and maybe using this exercise as an example.
$X$ and $Y$ are independent discrete stochastic variables, with probability functions
$$p_X(0) = P\{X=0\} = \frac 14; \quad p_X(1) = P\{X=1\} = \frac 34.$$
$$p_Y(0) = P\{Y=0\} = \frac 14; \quad p_Y(1) = P\{Y=1\} = \frac 34.$$
I need to specify the distribution of $X + Y$.
The probability space for X+Y = S = {0,1,2}, through the independence assumption we can calculate the event of $P(X = 0, Y=0) = P(X=0)*P(Y=0)$ therefore the following holds:
$P(X+Y = 0) = P(0 + 0 = 0) = P_x(0)*p_y(0) = 1/16$ $P(X+Y = 1) = P(1 + 0 = 1) | P(0 + 1 = 1) = P_x(1)*P_y(0) + P_x(0)*P_y(1) = 3/4*1/4 + 1/4*3/4 = 3/16 + 3/16 = 6/16 = 3/8$ $P(X+Y = 2) = P(1 + 1 = 2) = P_x(1)*P_y(1) = 3/4*3/4 = 9/16$
In general cases you can use the convolution sum (see, http://en.wikipedia.org/wiki/Convolution)