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Problem regarding E[Y^2]

$X$ is uniformly distributed in $[0,1]$, and $Y$ is uniformly distributed in $[0,X]$
(a) Find $\mathbb{E}(Y)$ and $\text{Var}(Y)$
Solution:
By conditioning on $X$, we obtain the first two orders of statistics of $Y$ as $\mathbb{E}(Y)=\mathbb{E}_X[\mathbb{E}_Y(Y|X)]=\mathbb{E}_X[\frac{X}{2}]=\frac14$
$\mathbb{E}(Y^2)=\mathbb{E}_X[\mathbb{E}_Y(Y^2|X)]=\mathbb{E}_X[\frac{X^2}{3}]=\frac19$
$\text{Var}(Y)= \mathbb{E}(Y^2)-\mathbb{E}^2(Y) = \frac19-(\frac14)^2=\frac{7}{144}$

So, I understand why E[Y]=1/4 without doing any integrals and stuff, but I'm a bit confused as to how to get the E[Y^2] value.

From where I understand you have to take the integral from 0 to x of y^2/x. But I don't understand where they got this integral from.

Glen_b
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useroe
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  • The variance of a random variable uniformly distributed on $(a,b)$, $X \sim \mathrm{Unif}(a,b)$ is given by $\mathrm{Var}(X) = (b-a)^2 / 12$. If you know $\mathbb{E}(X)^2$, how can you use this to determine $\mathbb{E}(X^2)$? – bashfuloctopus Sep 26 '17 at 00:18
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    Presumably "orders of statistics" is intended to mean (raw) "moments". It's a little confusing because it sounds like it's trying to say "order statistics" which is something else altogether. – Glen_b Sep 26 '17 at 03:49
  • @useroe If you found the answer useful, please accept it (by clicking the check / tick next to the answer ). – Mathemagical Oct 03 '17 at 02:40

1 Answers1

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The uniform distribution on $(0,X)$ has density $\frac{1}{X}$ $$\mathbb{E}_Y(Y^2 | X) = \int_0^X y^2 \frac{1}{X} dy = \frac{X^2}{3}$$

And $$\mathbb{E}_X \left[\frac{X^2}{3}\right] = \int_0^1 \frac{x^2}{3} dX = \frac{1}{9}$$

Mathemagical
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