How to calculate E[X] using conditioning. Let a random variable Y = X^2, While X is a random variable that can take the following values {-2,-1,0,1,2}. Each value can occur with equal probability i.e., 1/5.
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What's the question? If the goal is to compute $E[X]$, where does $Y$ come in? – lulu Dec 25 '19 at 12:45
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@lulu I think Y will be used as following formula could be used to find E[X]........"E[X]=E[E[X|Y]]" But I'm not so much sure about that. – Xzcess Dec 25 '19 at 13:52
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But $E[X]=0$, obviously. What's the problem? – lulu Dec 25 '19 at 13:54
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We have \begin{align} \mathbb E[X] &= \sum_{n=0}^2 \mathbb E[X\mid Y=n]\mathbb P(Y=n)\\ &=\mathbb E[X\mid Y=0]\cdot\frac15 + \mathbb E[X\mid Y=1]\cdot\frac25 + \mathbb E[X\mid Y=4]\cdot\frac25\\ &=0+0+0=0. \end{align}
Math1000
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@Math1000 isn't the probability for Y=4 is 2/5 ? Well the ans will still be zero but just for the correction. – Xzcess Dec 25 '19 at 20:41
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$$ E(X) = \left( \frac{1}{5} \right)(-2) + \left( \frac{1}{5} \right)(-1) + \left( \frac{1}{5} \right)(0) + \left( \frac{1}{5} \right)(1) + \left( \frac{1}{5} \right)(2) $$ $$ E(X) = \frac{ -2 -1 + 1 + 2} {5} $$ $$ E(X) = 0 $$ Now for the $E(Y)$. $$ E(Y) = \left( \frac{1}{5} \right)((-2)^2) + \left( \frac{1}{5} \right)((-1)^2) + \left( \frac{1}{5} \right)(0^2) + \left( \frac{1}{5} \right)(1^2) + \left( \frac{1}{5} \right)(2^2) $$ $$ E(Y) = \frac{ 4 + 1 + 0 + 1 + 4}{5} $$ $$ E(Y) = \frac{10}{5} = 2 $$
Bob
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This is just a simple method but I'm supposed to find E[X] by conditioning method. – Xzcess Dec 25 '19 at 13:48
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@Xzcess I have not heard of the conditioning method. Are you sure you have that term right? – Bob Dec 25 '19 at 13:59