Question

 Let X be a discrete random variable with the following PMF pX(k) = 0.2 for k = 0 0.2 for k = 1 0.3 for k = 2 0.3 for k = 3 0 otherwise (i) Let Y = X(X − 1)(X − 2), find pY (y). (ii) Find E[Y ] and V AR[Y ]. 

Answer 1

The random variable Y can take the following values:

0(0-1)(0-2)=0

1(1-1)(1-2)=0

2(2-1)(2-2)=0

3(3-1)(3-2)=6

Then

$P(Y = 0) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2 + 0.2 + 0.3 = 0.7$

$P(Y = 6) = P(X = 3) = 0.3$

i) We have a distribution series of the random variable Y

$\begin{matrix} Y&0&6\\ P&{0.7}&{0.3} \end{matrix}$

ii) Let's find

$E[Y] = \sum {{y_i}} {p_i} = 0 \cdot 0.7 + 6 \cdot 0.3 = 1.8$

$Var[Y] = E[{Y^2}] - {E^2}[Y] = 0 \cdot 0.7 + 36 \cdot 0.3 - {1.8^2} = {\rm{7}}{\rm{.56}}$