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Two random variables $x$, and $y$, with PDF $f_x(x)$ and $f_y(y)$.

Then a third variable $z = x + y + c$, with $c$ being a constant.

What is the probability of $x > x'$, given $z$? ($x'$ is a cutoff value of $x$.)

That is, what is $P(x > x' | z)$. Thank you so much.

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Also, suppose both x and y are normal distributions, what would be Prob (x > x'|z)? –  user79634 May 27 '13 at 4:31

1 Answer 1

Let me rewrite this:

Consider some independent random variables $X$ and $Y$, with density $f_X$ and $f_Y$ respectively, and $Z=X+Y+c$ for some constant $c$. Compute $P[X\gt x\mid Z]$, for every $x$.

There is no general answer to this question, but, when $X$ and $Y$ are both normal, one can go further.

Thus, assume without loss of generality that $X$ is centered normal vith variance $a^2$ and that $Y$ is centered normal vith variance $b^2$, then $(X,Z)$ is a centered gaussian vector with covariance matrix $\begin{pmatrix}a^2&a^2\\ a^2&a^2+b^2\end{pmatrix}$. In particular, $X=cZ+dU$, where $U$ is standard normal and independent of $Z$, for some suitable $(c,d)$. One can check that $c=a^2/(a^2+b^2)$ and $d=ab/\sqrt{a^2+b^2}$. Thus, the conditional distribution of $X$ conditionally on $Z$ is gaussian with mean $cZ$ and variance $d^2$, in particular, $$ P[X\gt x\mid Z]=P\left[U\gt\frac{x-cZ}d{\Large\mid} Z\right]=1-\Phi\left(\frac{x-cZ}d\right). $$

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