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Question
The random variable X can take only the values 0, 1, 2, 3. Give that P(X = 0) = P(X = 1) = p and P(X = 2) = P(X = 3) such that `Sigmap_i x_i^2 = 2Sigmap_ix_i`. Find the value of p
Solution
It is given that the random variable X can take only the values 0, 1, 2, 3.
Given:
P(X = 0) = P(X = 1) = p
P(X = 2) = P(X = 3)
Let P(X = 2) = P(X = 3) = q
Now,
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
`=> p + p + (q + q) = 1`
`=> q = (1- 2p)/2`
Since, `Sigmap_ix_i^2 = 2Sigmap_ix_i`
`=> 0 + p(1)^2 + ((1-2p)/2)(2^2 + 3^2) = 2[0 + p + ((1-2p)/2) (2 + 3)]`
`=> p + 13/2 (1- 2p) = 2[p + 5/2(1-2p)]`
`=> p + 13/2 - 13p = 2p + 5 - 10p`
`=> 13/2 - 12p = -8p + 5`
`=> 4p = 3/2`
`=> p = 3/8`
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