Winsorized Normal Variable Moments

Worked Solution

How to Think About It: You are clipping a normal distribution -- keeping the middle and zeroing out the tails. Since $X$ is symmetric about zero and the clipping is symmetric, $E[Y]$ should be zero by symmetry. The variance is the interesting part: it depends on how much of the distribution you keep. When $a$ is tiny, you zero out almost everything, so $\text{Var}(Y) \to 0$. When $a$ is huge, you keep almost everything, so $\text{Var}(Y) \to 1$. The formula for $\text{Var}(Y)$ in between requires computing $E[Y^2]$.

Quick Estimate: For $a = 2$, you keep about 95% of the distribution. The zeroed-out tails beyond $\pm 2$ contribute little to the variance of the original $N(0,1)$, so $\text{Var}(Y)$ should be close to but slightly less than 1. For $a = 0.5$, you keep only about 38% of the distribution -- most values get zeroed out, so $\text{Var}(Y)$ should be noticeably less than 1.

Approach: Use symmetry for $E[Y]$, then compute $E[Y^2]$ via integration.

Formal Solution:

Part 1: $E[Y]$

$E[Y] = \int_{-a}^{a} x \, \phi(x) \, dx = 0$

by symmetry, since $x \cdot \phi(x)$ is an odd function integrated over a symmetric interval.

Part 1: $\text{Var}(Y) = E[Y^2]$

Since $E[Y] = 0$, we have $\text{Var}(Y) = E[Y^2]$.

$E[Y^2] = \int_{-a}^{a} x^2 \phi(x) \, dx = 2 \int_0^a x^2 \phi(x) \, dx$

using that $x^2 \phi(x)$ is an even function. We evaluate this via integration by parts. Recall that $\phi'(x) = -x\phi(x)$, so $x\phi(x) = -\phi'(x)$.

$\int_0^a x^2 \phi(x) \, dx = \int_0^a x \cdot (-\phi'(x)) \, dx$

Integrate by parts with $u = x$, $dv = -\phi'(x)dx$:

$= \left[-x\phi(x)\right]_0^a + \int_0^a \phi(x) \, dx = -a\phi(a) + \Phi(a) - \Phi(0)$

$= -a\phi(a) + \Phi(a) - \frac{1}{2}$

Therefore:

$E[Y^2] = 2\left[-a\phi(a) + \Phi(a) - \frac{1}{2}\right] = -2a\phi(a) + 2\Phi(a) - 1$

So:

$\boxed{E[Y] = 0, \quad \text{Var}(Y) = 2\Phi(a) - 1 - 2a\phi(a)}$

Part 2: Limits

As $a \to 0^+$: $\Phi(a) \to \Phi(0) = 1/2$ and $a\phi(a) \to 0 \cdot \phi(0) = 0$. So:

$\lim_{a \to 0^+} \text{Var}(Y) = 2(1/2) - 1 - 0 = 0$

Interpretation: When $a \to 0$, the cutoff is so tight that essentially all values of $X$ get zeroed out, so $Y \approx 0$ almost surely. No variance remains.

As $a \to \infty$: $\Phi(a) \to 1$ and $a\phi(a) = a \cdot \frac{1}{\sqrt{2\pi}} e^{-a^2/2} \to 0$ (exponential decay beats linear growth). So:

$\lim_{a \to \infty} \text{Var}(Y) = 2(1) - 1 - 0 = 1$

Interpretation: When $a \to \infty$, the cutoff never activates -- you keep all of $X$. So $Y = X$ and $\text{Var}(Y) = \text{Var}(X) = 1$.

Answer: $E[Y] = 0$ by symmetry. $\text{Var}(Y) = 2\Phi(a) - 1 - 2a\phi(a)$. The variance increases monotonically from 0 (when $a \to 0$, everything is clipped) to 1 (when $a \to \infty$, nothing is clipped).