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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
Solution
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`.
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