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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).
Solution
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).
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