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Let $X$ and $Y$ be random variables such that $X \vert Y=y$ is normal distributed as $N(y,1)$ and Y is a continues random variable with PDF:

$f_Y(y)=3y^2$

for $0<y<1$

and $0$ otherwise.

Find $Var(X)$


My idea is to find the marginal PDF for X and from there it's easy to find the variance. Now since the conditional is normal distributed, I should be able to find the simulatanous PDF by multiplying the the conditional PDF by the marginal PDF for Y. However when I try to use integration to find the marginal PDF for X with limits 0 and infinity, I get an uncomputable integral.... Why is this happening and/or is there another easier method? Thanks in advance!

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  • $\begingroup$ You can find the variance from $E(X)=E\, [E(X\mid Y)]$ and $E(X^2)=E\, [E(X^2\mid Y)]$. $\endgroup$ – StubbornAtom Jan 8 '19 at 14:58
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Use law of total variance, $$\mathbb{Var}(X)=\mathbb{E}[\mathbb{Var}(X|Y)]+\mathbb{Var}(\mathbb{E}[X|Y])$$

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  • $\begingroup$ Okay so E[Var(XIY)] must be E[1]=1 since we know it's normal distributed with variance = 1. But how do I calculate Var(E[XIY])...? Is it using LOTUS? And wouldn't I also need to use Y's PDF in some way? $\endgroup$ – CruZ Jan 8 '19 at 16:54
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    $\begingroup$ @CruZ You are almost there. Think what random variable $\mathbb{E}[X|Y]$ is, and find its variance. $\endgroup$ – kludg Jan 8 '19 at 17:04
  • $\begingroup$ Okay so E[XIY]=y so I just have to calculate Var(Y)=E[Y^2]-(E[Y])^2? And I can do that using LOTUS! Is that correct? $\endgroup$ – CruZ Jan 8 '19 at 17:21
  • $\begingroup$ Looks reasonable to me, I would solve it so. $\endgroup$ – kludg Jan 8 '19 at 17:24

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