Question

The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P(X = 1) = p and that E(X²) = E[X], find the value of p

Answer 1

Given that: X = 0, 1, 2

And P(X) at X = 0 and 1 is p.

Let P(X) at X = 2 is x

⇒ p + p + x = 1

⇒ x = 1 – 2p

Now we have the following distributions.

X	0	1	2
P(X)	p	p	1 – 2p

∴ E(X) = 0.p + 1.p + 2(1 – 2p)

= p + 2 – 4p

= 2 – 3p

And E(X²) = 0.p + 1.p + 4(1 – 2p)

= p + 4 – 8p

= 4 – 7p

Given that: E(X²) = E(X)

∴ 4 – 7p = 2 – 3p

⇒ 4p = 2

⇒ p = `1/2`

Hence, the required value of p is `1/2`.