正对称分布下$E\left[\frac{1}{X}\right]$的下界推导问题

Great question! You already nailed the first part with Jensen's inequality, which leverages the convexity of (1/x) (for (x>0)) to get the basic lower bound. To add the variance term, we’ll combine the distribution’s symmetry with a second-order look at the function (1/x). Here’s how to do it:

Step 1: Pair symmetric points using the distribution’s symmetry #

Since (X) is symmetric around its mean (m), every value (m+t) that (X) can take has a corresponding value (m-t) with equal probability. Let (T = |X - m|); we can rewrite the expectation as:
[
E\left[\frac{1}{X}\right] = \frac{1}{2}E\left[\frac{1}{m+T} + \frac{1}{m-T}\right]
]

Step 2: Simplify the sum of reciprocals for symmetric pairs #

First, compute the sum inside the expectation:
[
\frac{1}{m+t} + \frac{1}{m-t} = \frac{(m-t) + (m+t)}{(m+t)(m-t)} = \frac{2m}{m^2 - t^2}
]
Next, use the geometric series expansion for (1/(1 - t^2/m2)) (valid because (X>0) means (t < m), so (t^2/m2 < 1)):
[
\frac{2m}{m^2 - t^2} = \frac{2}{m} \cdot \frac{1}{1 - t^2/m2} = \frac{2}{m}\left(1 + \frac{t^2}{m2} + \frac{t^4}{m4} + \dots\right)
]
All higher-order terms ((t^4/m4) and beyond) are non-negative, so we can truncate the series to get a lower bound:
[
\frac{1}{m+t} + \frac{1}{m-t} \geq \frac{2}{m} + \frac{2t^2}{m3}
]

Step 3: Take the expectation to get the final inequality #

Apply expectation to both sides. Remember that (E[T^2] = E[(X - m)^2] = \sigma^2):
[
E\left[\frac{1}{m+T} + \frac{1}{m-T}\right] \geq E\left[\frac{2}{m} + \frac{2T^2}{m3}\right] = \frac{2}{m} + \frac{2\sigma^2}{m3}
]
Multiply both sides by (1/2) to get back to the original expectation:
[
E\left[\frac{1}{X}\right] \geq \frac{1}{m} + \frac{\sigma^2}{m3}
]

Quick note on why symmetry matters #

If you tried to use a second-order Taylor bound directly for individual (x), you’d run into a problem: (1/x)’s second derivative decreases with (x), so the Taylor lower bound fails for (x > m) (e.g., (1/(m+t) < 1/m - t/m^2 + t^2/m3) when (t>0)). Symmetry fixes this by pairing each (m+t) with (m-t)—their combined sum of reciprocals ends up being greater than the combined Taylor bounds, and the problematic linear terms cancel out entirely.

内容的提问来源于stack exchange，提问作者Ethan