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A random variable $N$ is uniformly distributed on $\{1,2,...,10\}$ Let $X$ be the indicator of the event($N\le 5$) and $Y$ be the indicator of the event ($N$ is even)

So I have to find $E(X)$ and $E(Y)$

From the formula sheet I know

$$E(X) = \sum_i x_iP(x_i)$$ $$E(Y) = E[E(Y\mid X=x_i)] $$

but i don't get how to select the xi and P(xi) to solve this problem. The answer says $E(X)=E(Y) =1/2$ but It does not have any process inside so I could not understand can anyone help me about this

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Looking at RHS of equality: $$\mathbb EX=\sum_ix_iP(x_i)$$ we observe that a term $x_iP(x_i)$ is only relevant if: $$x_iP(x_i)\neq0$$

That will be the case here if and only if $x_i=1$ and in that case: $$P(x_i)=P(1)=\Pr(X=1)=\Pr(N\leq5)=\frac5{10}=\frac12$$

So we find:$$\mathbb EX=1\cdot\frac12=\frac12$$You can find $\mathbb EY$ on a similar way.

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  • $\begingroup$ why there will be only case xi = 1? $\endgroup$ – 헬창공돌이 Apr 17 at 10:59
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    $\begingroup$ @헬창공돌이 Because $X$ is an indicator, then by definition it equals 1 when $N\leq 5$ and $0$ otherwise.$$X=\begin{cases}1&:& N\leq 5\\0&:& N>5\end{cases}$$ So therefore: $\mathsf E(X)=0\cdot\mathsf P(N>5)+ 1\cdot\mathsf P(N\leq 5)$ . $\endgroup$ – Graham Kemp Apr 17 at 11:02
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The two rv's are both bernulli with parameter 0.5 thus their expectation is $0\times 0.5 + 1\times 0.5 =0.5$

This because the rv that describes the event $N\leq 5$ can take only the values 0 and 1 (false or true) with probability 0.5 as you have 5 favourable events among 10 equiprobable ones

Similar reasoning for the other rv

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  • $\begingroup$ could you explain more further sorry T_T $\endgroup$ – 헬창공돌이 Apr 17 at 9:22
  • $\begingroup$ Added some details in my answer. If it has been useful you can mark it as accepted $\endgroup$ – tommik Apr 17 at 9:36

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