Expectation of inverse of normal RV, given that it is below a certain value

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?

• this link looks relevant. stats.stackexchange.com/questions/70045/… – mlofton Aug 14 at 4:56
• As with the link above, the expectation will not converge, given the 1/X and the domain of support including 0. – wolfies Aug 14 at 6:52
• Because this conditional variable has positive continuous density in a neighborhood of zero, stats.stackexchange.com/questions/299722 demonstrates the expectation does not exist. – whuber Aug 14 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 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 at 18:57

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)
 998638
a = mean(1/xc); a
 0.04174508
hist(1/xc, prob=T, col="skyblue2")
abline(v=a, col="red") Tangentially related application: Here

• $X>T$ (for T positive) is not the same as $X < T$, when considering $\frac1X$ – wolfies Aug 14 at 7:07
• Right: Should have noted explicitly that I was changing the rules to get an expectation that exists. Editing. – BruceET Aug 14 at 7:16