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

Question

Philoid · Accepted Answer

Given that-

X=0,1,2 and P(X) at X=0 and 1,

Let at X=2, P(X) is x.

p+p+x=1

x=1-2p

We get the following distribution

∴E(X)= XP(X)

= 0×P + 1×P+ 2(1-2P)

= P+2-4P = 2-3P

And E(X)²= X² P(X)

= 0×P+ 1× P+ 4×(1-2P)

= P+4-8P=4-7P

Also, given that E(X²) =E(X)

4-7p= 2-3p

4p=2