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So if you have a distribution fx(x) and you observed x=a, how do you calculate E{x | x=a} = integral (x*fx(x|x=a) from -inf to inf, or more specifically, fx(x|x=a)

I am having trouble when using Bayes or finding Fx(x|x=a) because I end up with P(x=a) which is just a constant. Is there a simple example somewhere? Searching only leads me to bivariate distributions.

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up vote 3 down vote accepted

Unless I misunderstood your problem: your conditional density is just a Dirac delta function in $x=a$:

$$f_x(x|x=a) =\delta(x-a)$$

Hence, $E(x | x=a) = a$ ... as intuition says.

If you want to apply the definition, you can use a limiting procedure (because, for a continuous variable the event $X=a$ has actually zero probability), considering instead the event that $X$ takes a value around $a$, in a neighborhood of length $dx$. But that's not necessary. You already know that the joint probability (the one that goes into the numerator) is zero if $x\ne a$. Hence, as the conditional density must integrate to one, it must be a dirac delta function.

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Please check your post. – AD. Oct 6 '10 at 18:07
Thanks for the answer and explanation, makes sense to me. – Eruditass Oct 6 '10 at 18:47

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