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$$Y|X=x \sim N(x,1)\\X\sim N(\mu,\sigma^2 )$$ What distribution does $X|Y=y$ follow?

My initial startegy was to $f_{Y|X}f_X=f_{X,Y}$ and solve for $f_{X|Y}=f_{X,Y}/f_{Y}$ . Computing for $f_{X,Y}$, I get the following: $$f_{X,Y}=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{(y-x)^2}{2}\right\}$$ And in trying to compute for $f_{Y}$, I was trying to integrate above w.r.t $x$, but I was stuck. I am not sure if that's integratable, and if this is a right approach to solve this question. I am curious if there is some kind of tricks/insights I am missing.

Alecos Papadopoulos
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Alby
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    Generally, the gist of these kind of problems go: combine and expand the exponents, collect like terms, complete the square, write quadratic as $(x-g(y,\mu,...))^2+S$, spot the density. – Glen_b Oct 15 '13 at 23:06
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    You can now find a shorter way to calculate your integral, here: http://stats.stackexchange.com/questions/73157/calculation-of-an-unconstrained-normal-distribution-starting-from-a-censored/73327#73327 – Alecos Papadopoulos Oct 20 '13 at 23:41

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Separate the exponents to terms that do not contain $x$ and those that contain $x$ . You will obtain an integrand that can be written in the form $e^{-ax^2-bx}$ (all other terms go out of the integral, since you want to integrate w.r.t $x$). Then Gradshteyn & Ryzhik (2007), "Table of Integrals, Series and Products", 7th ed. p.336, eq. 3.322(2) give the formula:

$$\int_{0}^{\infty}\exp\left\{−\frac {x^2}{4\beta}−\gamma x\right\}dx = \sqrt {\pi\beta} \exp\left\{\beta \gamma^2\right\} \left[1-\operatorname{erf}(\gamma \sqrt \beta)\right]$$

I presume you can turn a $\int_{-\infty}^{\infty}$ integral into $\int_{0}^{\infty}$ integrals.

It's going to be a bit long and tedious, easy to make an algebraic mistake. After you're done, don't forget to remember the connection of the error function with the cdf of the standard normal.

Alecos Papadopoulos
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  • For a more general case: $Y|X=x \sim N(x,\sigma_x^2)$ and $X\sim N(\mu_x,\sigma^2_x)$ I've $\beta = \frac{2\sigma_x^2 + 2\sigma_y^2}{\sigma_y^2\sigma_x^2}$ and $\gamma = \frac{-4\sigma_x^2 y - 4\sigma^2_y\mu_x}{4\sigma_y^2\sigma_x^2}$. It seems that $\int_{-\infty}^0 \text{exp}{-\frac{x^2}{4\beta}-\gamma x}dx = \int_{0}^\infty \text{exp}{-\frac{x^2}{4\beta}-\gamma x}dx$. With this I'll get smth. like a density multiplied by the $\text{erf}$. But since $\text{erf}$ integrates over $\gamma\sqrt{\beta}$ which depends upon $y$ how am I supposed to find the expected value and variance of $y$? – Druss2k Oct 20 '13 at 15:03
  • Short: $\text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x \text{exp}{-\tau^2}d\tau$. Because $\tau = \gamma\sqrt{\beta} = \text{f}(y,\mu_x,\sigma_x^2,\sigma_y^2)$ the $\text{erf}$ part seems rather complicated. – Druss2k Oct 20 '13 at 15:13
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    $\int_{-\infty}^0 \text{exp}{-\frac{x^2}{4\beta}-\gamma x}dx = \int_{0}^\infty \text{exp}{-\frac{x^2}{4\beta}+\gamma x}dx$. Use the relation of erf with the standard normal cdf, and we'll take it from there. – Alecos Papadopoulos Oct 20 '13 at 17:19