# Transformations of Discrete Random Variables and Vectors

In the discrete case, the question Given a random variable or vector X and a Borel measure function or mapping g(x), how is the distribution of Y = g(X) related to the distribution of X? is easy to answer. If P[X є {хь x2,…}] = 1 and

g(x1), g(x2),… are all different, the answer is trivial: P(Y = g(xj)) = P(X = Xj). If some of the values g(x1), g(x2),… are the same, let {y1, y2,…} be the set of distinct values of g(x1), g(x2),… Then

TO

P(Y = yj) = £ I[y = g(Xi)]P(X = xi). (4.13)

i=1

It is easy to see that (4.13) carries over to the multivariate discrete case.

For example, if X is Poisson(X)-distributed and g(x) = sin2(nx) = (sin(nx))2 – and thus for m = 0, 1, 2, 3,…, g(2m) = sin2(nm) = 0 and g(2m + 1) = sin2(nm + n/2) = 1 – then P(Y = 0) = e-XY°TO=0 xlj/(2j)! and P(Y = 1) = e-kJ2j=0 X2j+1 /(2j + 1)!

As an application, let X = (X 1, X2)T, where X1 and X2 are independent Poisson(X) distributed, and let Y = X1 + X2. Then for y = 0, 1,2,…

TO TO

P(Y = y) = EE I[y = i + j]P(X1 = i, X2 = j)

i = 0 j = 0

(2X)y

= exp(-2X)^-. (4.14)

y!

Hence, Y is Poisson(2X) distributed. More generally, we have

Theorem 4.1: If for j = 1,…, k the random variables Xj are independent Poisson(X j) distributed, thenY^j=1 Xj is Poisson Qfkj=1 X j) distributed.

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