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Discrete random variable $\Theta$ is uniformly distributed between 1 and 100. Given $\Theta$ discrete random variable $X$ is uniformly distributed between 1 and $\Theta$. Show that

$E[X^2] = \frac{1}{100}\sum\limits_{x=1}^{100} x^2 \bigg(\sum\limits_{\theta=x}^{100} \frac{1}{\theta}\bigg)$

How to use $E[E[X^2|\Theta]]=E[X^2]$ to get the answer. Thank you

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  • $\begingroup$ So what have you tried so far? For example, what is $E[X^2 \mid \Theta=\theta]$? $\endgroup$
    – Henry
    Commented Aug 17, 2023 at 18:06
  • $\begingroup$ It should be $\sum\limits_{\theta=x}^{100} x^2 \frac{1}{\theta}$. In my opinion $\endgroup$
    – Raja Ali Riaz
    Commented Aug 17, 2023 at 18:14
  • $\begingroup$ No - it should be $\sum\limits_{x=1} ^ \theta x^2 \frac1\theta$. Can you find what that is? Then try $\sum\limits_{\theta=1} ^ {100} \left(\sum\limits_{x=1} ^ \theta x^2 \frac1\theta\right) \frac1{100}$ $\endgroup$
    – Henry
    Commented Aug 17, 2023 at 18:20
  • $\begingroup$ Agreed for the first limit but the second equation is not clear. We are taking expectation of the first equation sum is ranging for $\theta$ with it's pmf. How to reduce it to answer. $\endgroup$
    – Raja Ali Riaz
    Commented Aug 17, 2023 at 18:29
  • $\begingroup$ Thank you Henry got it Thank you $\endgroup$
    – Raja Ali Riaz
    Commented Aug 17, 2023 at 19:13

1 Answer 1

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If $X$ and $Z$ are discrete, then $$ E(X) = E(E(X \mid Z)) \mathop{=}^{\mathrm{def}} \sum_z E(X \mid Z = z) P(Z = z). $$ If you know $E(X \mid Z = z) = u(z),$ then $$ E(X) = \sum_z u(z) P(Z = z). $$ Now apply this result replacing the appropriate values.

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