How to compute \begin{align} P(W>0| X+W=u) \end{align} where $X$ and $W$ are independent standard normal.
I am asked to do this via Baye' Rule.
How to compute $P(W>0| X+W=u)$
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$\begingroup$
probability
probability-theory
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1@MeesdeVries I am not sure how to use both probabilities and probability density function is Baye's rule. For example, is $P(W>0|X+W=u)=\frac{P(W>0) f_{X+W|W}(u|W>0)}{f_{X+W}(u)}$ correct? I don't think this is. – 2017-02-07
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3For a solution with (almost) no computation, note that $$U=\frac{W+X}{\sqrt 2}\qquad V=\frac{X-W}{\sqrt 2}$$ defines a standard i.i.d. random vector $(U,V)$ hence one is after $$P(U>V\mid U=\sqrt2 u)=P(V<\sqrt2 u\mid U=\sqrt2 u)=P(V<\sqrt2 u)=\Phi(\sqrt2 u)$$ – 2017-02-07
1 Answers
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Hint: First try to compute for $y \in \mathbb{R}$ $$f_{W|X+W}(y|u)$$ Then the result you desire will be obtained by integrating wrt $y$ from $0$ to $\infty$. Let me know if you need additional help.
Also $f_{X+W|W}(y|u)=f_{X}(y-u)$ as $X$ and $W$ are independent.