since there are only countably many non-zero contributions to this sum. Thus, if Y = aX +b with a 6= 0, then P(Y = y) = P(aX + b = y) = P

????

X = a−1(y − b)

for y ∈ R, while if Y = X2, then P(Y = y) =





P

????

X = √y

+ P

????

X = −√y

if y > 0, P(X = 0) if y = 0, 0 if y < 0.