# Conditional Distributions and Probabilities

Suppose that $Y=A+\epsilon$ where $\epsilon$ is a RV and given some other random variable $\eta$ we have that:

$\epsilon|\eta$ ~ $N(\rho\eta,\sigma^2)$

Suppose I was asked to find $Pr(Y=y|\eta)$ This (I think) is easily equal to: $\frac{1}{\sigma}\phi(\frac{y-A-\rho\eta}{\sigma})$ where $\phi$ is the standard normal pdf.

Now I suppose I need to find $Pr(Y=y|\eta>x)$ The answer I have for this, using conditional probability formula (P(A|B)=$\frac{P(A\cap B)}{P(B)})$is:

$\frac{\int_x^{\infty}\frac{1}{\sigma}\phi(\frac{y-A-\rho z}{\sigma})dz}{P( \eta>x)}=\frac{\Phi(\frac{y-A-\rho x}{\sigma})}{(P( \eta>x))\rho}$

which does not seem right to me.

Edit: I forgot about the y's

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$Y$ is a continuous random variable. Isn't it true that $Pr(Y=y|\eta)=0$? –  Patrick Li Jan 30 '13 at 21:31
It wasn't the best choice of notation. I meant I wanted the PDF. –  Greg Jan 30 '13 at 22:09

Obviously, the answer should depend on the distribution of $\eta$. Assume that $\eta$ has PDF $g$ and CDF $G$, then the conditional PDF $f_x$ of $Y$ conditionally on $\eta\gt x$ is such that $$f_x(y)=\frac1{1-G(x)}\int_x^{+\infty}\frac1\sigma\phi\left(\frac{y-A-\rho t}{\sigma}\right)\,g(t)\mathrm dt,$$ where $\phi$ is the standard gaussian PDF and $$1-G(x)=\mathbb P(\eta\gt x)=\int_x^{+\infty}g(t)\mathrm dt.$$

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Since $Y= A+ \epsilon$, then we know that $\epsilon = Y-A$. Also $P(Y = y) = P(A+\epsilon = y) \Longleftrightarrow P(\epsilon = y-A)$. So it seems that $$P(Y = y| \eta) = P(\epsilon = y-A | \eta)$$ $$= \Phi \left(\frac{y-A-\rho \eta}{\sigma} \right)$$

So $$P(Y = y| \eta > x) = P(\epsilon = y-A| \eta>x)$$

$$= \int_{x}^{\infty} \Phi \left(\frac{y-A-\rho \eta}{\sigma} \right) \ d \eta$$

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Thanks for pointing that out, I've made the corrections. However, the main question remains. –  Greg Jan 30 '13 at 20:23
@Greg: Since we are given the distribution $\epsilon| \eta \sim N(\rho \eta, \sigma^2)$, I don't think we need the extra $\frac{1}{\sigma}$. –  proton Jan 30 '13 at 20:42
Careful: $P(Y=y)=0$ and $P(Y=x\mid \eta\gt x)=0$ for every $y$. –  Did Jan 30 '13 at 22:03