Question

Two probability distributions of the discrete random variable X and Y are given below.

X	0	1	2	3
P(X)	`1/5`	`2/5`	`1/5`	`1/5`

 

Y	0	1	2	3
P(Y)	`1/5`	`3/10`	`2/10`	`1/10`

Prove that E(Y²) = 2E(X).

Answer 1

First probability distribution is given by

X	0	1	2	3
P(X)	`1/5`	`2/5`	`1/5`	`1/5`

We know that, E(X) = `sum_("i" = 11)^"n" "P"_"i""X"_"i"`

⇒ E(X) = `0 * 1/5 + 1 * 2/5 + 2 * 1/5 + 3 * 1/5`

= `0 + 2/5 + 2/5 + 3/5`

= `7/5`

For the second probability distribution,

Y	0	1	2	3
P(Y)	`1/5`	`3/10`	`2/10`	`1/10`

E(Y²) = `0 * 1/5 + 1 * 3/10 + 4 * 2/5 + 9 * 1/10`

= `0 + 3/10 + 8/5 + 9/10`

= `28/10`

= `14/5`

Now E(Y²) = `14/5` and 2E(X) = `2*7/5 = 14/5`

Hence, E(Y²) = 2E(X).