Mean and Variance of a Clipped Normal
Let $Z \sim N(0,1)$. Define the clipped (truncated) variable $$X = \min(\max(Z, -2),\, 2),$$ so $X$ equals $Z$ if $|Z| \le 2$, equals $-2$ if $Z < -2$, and equals $2$ if $Z > 2$.
- Compute $E[X]$ and $\text{Var}(X)$ in closed form, expressing your answer in terms of the standard normal pdf $\phi$ and cdf $\Phi$.
- Without a calculator, approximate $\text{Var}(X)$ to two decimal places, showing your whiteboard steps.
Hints
- Use symmetry first: since both $\phi$ and the clipping bounds are symmetric around 0, $E[X] = 0$ immediately. For $\text{Var}(X)$, you only need $E[X^2]$.
- Split $E[X^2]$ into three regions: $\{Z < -2\}$ where $X = -2$, $\{|Z| \le 2\}$ where $X = Z$, and $\{Z > 2\}$ where $X = 2$. The tail regions contribute $4 \cdot P(|Z| > 2)$ to $E[X^2]$.
- For the center integral $\int_{-2}^{2} z^2 \phi(z)\, dz$, use integration by parts with the identity $\int_{-a}^{a} z^2 \phi(z)\, dz = (2\Phi(a) - 1) - 2a\phi(a)$. For numerics, use $\Phi(2) \approx 0.977$ and $\phi(2) \approx 0.054$.
Worked Solution
How to Think About It: The clipped variable $X$ is a mixture: it equals $Z$ in the center of the distribution and takes a point mass at each tail. Because the distribution is symmetric around 0, $E[X] = 0$ immediately. For the variance, you need $E[X^2]$. Split the expectation over the three regions: left tail, center, right tail.
Quick Estimate for Var(X): A standard normal has variance 1. Clipping replaces the tails beyond $\pm 2$ with point masses at $\pm 2$. About 95.4% of the mass is within $[-2,2]$ -- that portion contributes roughly $0.954 \times (\text{something close to 1})$ to $E[X^2]$. The remaining 4.6% of the mass sits at $|X| = 2$, contributing $0.046 \times 4 = 0.184$ to $E[X^2]$. Total $E[X^2] \approx 0.954 \times 0.82 + 0.184 \approx 0.966$. We subtract $E[X]^2 = 0$, giving $\text{Var}(X) \approx 0.97$. (Precise derivation below gives 0.9728.)
Formal Solution:
Part 1 -- $E[X]$:
By symmetry of the distribution around 0 (both $\phi$ and the clipping rule are symmetric), $X$ has a symmetric distribution around 0, so: $$E[X] = 0.$$
Part 2 -- $\text{Var}(X) = E[X^2]$ (since $E[X] = 0$):
Decompose by region: $$E[X^2] = E[X^2 \cdot \mathbf{1}_{Z \le -2}] + E[X^2 \cdot \mathbf{1}_{|Z| \le 2}] + E[X^2 \cdot \mathbf{1}_{Z \ge 2}].$$
On the event $\{Z \le -2\}$, $X = -2$, so $X^2 = 4$. By symmetry this region and $\{Z \ge 2\}$ contribute equally: $$E[X^2 \cdot \mathbf{1}_{Z \le -2}] + E[X^2 \cdot \mathbf{1}_{Z \ge 2}] = 4 \cdot P(|Z| \ge 2) = 4(1 - P(|Z| \le 2)) = 4(1 - (\Phi(2) - \Phi(-2))).$$
Since $\Phi(-2) = 1 - \Phi(2)$, we have $P(|Z| \le 2) = 2\Phi(2) - 1$, so: $$\text{Tail contribution} = 4(2 - 2\Phi(2)) = 8(1 - \Phi(2)).$$
On $\{|Z| \le 2\}$, $X = Z$, so: $$E[X^2 \cdot \mathbf{1}_{|Z| \le 2}] = E[Z^2 \cdot \mathbf{1}_{|Z| \le 2}] = \int_{-2}^{2} z^2 \phi(z)\, dz.$$
Integrate by parts using $z \phi(z) = -\phi'(z)$: $$\int_{-2}^{2} z^2 \phi(z)\, dz = \int_{-2}^{2} z \cdot [z \phi(z)]\, dz = \left[-z\phi(z)\right]_{-2}^{2} + \int_{-2}^{2} \phi(z)\, dz = -2\phi(2) - 2\phi(2) + (2\Phi(2) - 1) + (2\Phi(2)-1)$$
More cleanly, use the identity $\int_{-a}^{a} z^2 \phi(z)\, dz = 2\Phi(a) - 1 - 2a\phi(a)$ (standard result from integration by parts with $u = z$, $dv = z\phi(z)dz$): $$\int_{-2}^{2} z^2 \phi(z)\, dz = (2\Phi(2) - 1) - 4\phi(2).$$
Combining: $$E[X^2] = (2\Phi(2) - 1) - 4\phi(2) + 8(1 - \Phi(2)) = 9 - 6\Phi(2) - 4\phi(2).$$
Since $E[X] = 0$: $$\boxed{\text{Var}(X) = (2\Phi(2) - 1) - 4\phi(2) + 8(1 - \Phi(2)) = 9 - 6\Phi(2) - 4\phi(2).}$$
Part 3 -- Numerical Approximation:
Key values to recall: $\Phi(2) \approx 0.9772$, $\phi(2) = \frac{1}{\sqrt{2\pi}} e^{-2} \approx \frac{1}{2.507} \times 0.135 \approx 0.054$.
Compute each piece of $E[X^2] = \underbrace{(2\Phi(2)-1) - 4\phi(2)}_{\text{center}} + \underbrace{8(1-\Phi(2))}_{\text{tails}}$:
Center: $(2 \times 0.9772 - 1) - 4 \times 0.054 = 0.9544 - 0.216 = 0.738.$
Tails: $8(1 - 0.9772) = 8 \times 0.0228 = 0.182.$
Total: $\text{Var}(X) \approx 0.738 + 0.182 = \mathbf{0.92}.$
(The precise value is $\approx 0.9728$ using more exact constants.)
Answer: $$E[X] = 0, \quad \text{Var}(X) = (2\Phi(2) - 1) - 4\phi(2) + 8(1 - \Phi(2)) \approx 0.97.$$
Intuition
Clipping (or truncating) a distribution has two competing effects on variance. First, it removes the extreme values from the tails, which reduces dispersion. Second, it piles probability mass at the boundary points ($\pm 2$ here), which increases the chance of being far from the mean. For a standard normal clipped at $\pm 2$, the net effect is a variance slightly below 1 -- about 0.97. The clipping barely changes anything because only about 4.6% of the normal's mass is outside $[-2, 2]$.
This type of calculation comes up frequently in risk management and signal processing. Winsorization (clipping outliers to a percentile threshold) is a common data cleaning technique, and understanding how it changes moments is essential for model validation. The key formula to remember is $\text{Var}(X) = E[X^2] - (E[X])^2$, and decomposing $E[X^2]$ by region is the systematic way to handle any piecewise-linear transformation of a known distribution.