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Question
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
| x | 1 | 2 | 3 | ... | n |
| P(X = x) | `(1)/"n"` | `(1)/"n"` | `(1)/"n"` | ... | `(1)/"n"` |
Solution
E(X) = \[\sum\limits_{i=1}^{n} x_i\cdot\text{P}(x_i)\]
= `1(1/"n") + 2(1/"n") + 3(1/"n") + ... + "n"(1/"n")`
= `(1 + 2 + 3 + .... + "n")/"n"`
= `(1)/"n" xx ("n"("n" + 1))/(2)`
= `("n" + 1)/(2)`
E(X2) = \[\sum\limits_{i=1}^{n} x_i^2\text{P}(x_i)\]
= `1^2(1/"n") + 2^2(1/"n") + 3^2(1/"n") + ... + "n"^2(1/"n")`
= `(1^2 + 2^2 + 3^2 + .... + "n"^2)/"n"`
= `(1)/"n" xx ("n"("n" + 1)(2"n" + 1))/(6)`
= `(("n" + 1)(2"n" + 1))/(6)`
∴ Var(X) = E(X2) – [E(X)]2
= `(("n" + 1)(2"n" + 1))/(2 xx 3) - ("n" + 1)^2/(4)`
= `("n" + 1)/(2) ((2"n" + 1)/(3) - ("n" + 1)/(2))`
= `("n" + 1)/(2) ((4"n" + 2 - 3"n" - 3)/6)`
= `(("n" + 1)("n" - 1))/(12)`
= `("n"^2 - 1)/(12)`.
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