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Let the joint distribution of $x$ and $y$ has been given as:

$f(x,y)=\frac{e^{\frac{-y^2}{2}}}{y \sqrt{2\pi}}$ where $0\leq|x|\leq y \ <\infty$. It is required to obtain the mean and variance of the random variable $X$.

Although, I have solved these types of problems but in this one I am having trouble beacuse the expression which is coming out while obtaining the marginal density is not in nice and analytic for. Please help me with the problem. Thanks in advance.

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2 Answers 2

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You have probably figured this out by now.

In any case, the joint density factors as $$f(x,y)=\underbrace{\frac1{2y}\mathbf1_{|x|<y}}_{f_{X\mid Y}(x\mid y)}\cdot\underbrace{\sqrt{\frac2{\pi}}e^{-y^2/2}\mathbf1_{y>0}}_{f_Y(y)}$$

So $X$ conditioned on $Y=y$ is uniform on $(-y,y)$ for $y>0$ and $Y$ itself has a folded normal distribution. This is enough information to find mean and variance of $X$ using the law of total expectation/variance.

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It seems kind of hard. I tried this:

Using u-substitution, let $t = \frac{y^2}{2}$ and change the bound,

$$f(x) = \int_{|x|}^\infty \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}y} dy = \int_{\frac{x^2}{2}}^\infty \frac{e^{-t}}{2t \sqrt{2\pi}}dt$$

I checked the integral calculator and it seems that there is no definite integral too, but the result is the incomplete Gamma.

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