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I have a normal random variable $X$ with mean $\mu$ and variance $\sigma^2$. Any advice on how to compute the conditional expectation $E[\frac{1}{X}|X \leq T]$ where $T$ is a positive constant?

dotpixel
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    this link looks relevant. https://stats.stackexchange.com/questions/70045/mean-and-variance-of-inverse-of-a-normal-rv – mlofton Aug 14 '19 at 04:56
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    As with the link above, the expectation will not converge, given the 1/X and the domain of support including 0. – wolfies Aug 14 '19 at 06:52
  • Because this conditional variable has positive continuous density in a neighborhood of zero, https://stats.stackexchange.com/questions/299722 demonstrates the expectation does not exist. – whuber Aug 14 '19 at 13:02
  • @whuber Yes, I saw some of those links before, but if the mean is sufficiently large, would that make it okay? Would simulation be the best way to approximate this? – dotpixel Aug 14 '19 at 16:49
  • No, it doesn't help. My analysis (in one of those answers) shows the result has nothing to do with the mean. It's all about the fact that there's sufficient probability for $X$ to be close enough to zero that no expectation can exist. This will be true of any Normal distribution (although, to be sure, as a *practical* matter that probability may be negligible: but that's a different question). – whuber Aug 14 '19 at 18:57

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Comment: Simulation for $T = 10,$ which avoids taking reciprocals of values anywhere near $0.$ Then $E(\frac 1 X\, |\, X > 10) \approx 0.042.$ (As @Wolfies comments, this is a different problem.)

set.seed(2019)
x = rnorm(10^6, 25, 5)
xc = x[x > 10]
length(xc)
[1] 998638
a = mean(1/xc); a
[1] 0.04174508
hist(1/xc, prob=T, col="skyblue2")
 abline(v=a, col="red")

enter image description here

Tangentially related application: Here

BruceET
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