Moments of min of a random variable and a constant.

Question

Let $X$ be a random variables with Normal distribution: $N[m,\sigma^2]$. Let $\eta$ be a constant. Now, let $M=\min(X,\eta)$. What is the expectation and variance of $M$?

This question seems related but no one answered that quesion: Expectation of $\min(X, c)$ for $X$ truncated r.v. and $c$ constant

This question also seems related but it talks same about two or more random varible which is uniformly distributed: Expectation of Minimum of $n$ i.i.d. uniform random variables.

I tried to explicitly calculate the probability distribution of $M$, but it turned out too complicated.

Here is my current attempt; looking forward for your comments.)

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$E[M]=E[min(X,\eta)] =min(E[X],\eta) = min(m,\eta)$

$var[M]=var[min(X,\eta)]= min(var[X],\eta^2)= min(\sigma^2,\eta^2)$

Answer 1

Let $z_{\eta}\equiv(\eta-m)/\sigma$. Then

\begin{align} \mathbb{E}[X\wedge \eta]=&E[X1\{X\le \eta\}]+\eta \mathbb{E}[1\{X>\eta\}] \\ =&m\Phi(z_{\eta})-\sigma\phi(z_{\eta})+\eta(1-\Phi(z_{\eta})), \end{align}

\begin{align} \mathbb{E}[(X\wedge \eta)^2]=&\mathbb{E}[X^21\{X\le \eta\}]+\eta^2 \mathbb{E}[1\{X>\eta\}] \\ =&(m^2+\sigma^2)\Phi(z_{\eta})-\sigma(m+\eta)\phi(z_{\eta})+\eta^2(1-\Phi(z_{\eta})), \end{align}

and

$$ Var(X\wedge \eta)=\mathbb{E}[(X\wedge \eta)^2]-(\mathbb{E}[X\wedge \eta])^2. $$

Answer 2

Directly, $$\operatorname{E}[(\min(X, \eta))^k] = \int_{x=-\infty}^\eta x^k f_X(x) \, dx + \int_{x=\eta}^\infty \eta^k f_X(x) \, dx.$$ The second term is simply $\eta^k (1 - F_X(\eta)) = \eta^k \left( 1 - \Phi(\frac{\eta - \mu}{\sigma})\right)$. The first term is more challenging. For $k = 1$, we write $$\int_{x=-\infty}^\eta x f_X(x) \, dx = \mu F_X(\eta) + \sigma^2 \int_{x=-\infty}^\eta \frac{x-\mu}{\sigma^2} f_X(x) \, dx$$ and with the substitution $$z = \frac{(x-\mu)^2}{2\sigma^2}, \quad dz = \frac{x-\mu}{\sigma^2} \, dx,$$ we get $$\int_{x=-\infty}^\eta x f_X(x) \, dx = \mu F_X(\eta) + \sigma^2 \int_{z=\infty}^{(\eta - \mu)^2/(2\sigma^2)} \frac{e^{-z}}{\sqrt{2\pi}\sigma} \, dz = \mu F_X(\eta) - \frac{\sigma}{\sqrt{2\pi}} e^{-(\eta-\mu)^2/(2\sigma^2)}.$$ Thus $$\operatorname{E}[\min(X , \eta)] = \eta + (\mu-\eta) \, \Phi\!\left(\tfrac{\eta - \mu}{\sigma}\right) - \frac{\sigma}{\sqrt{2\pi}} e^{-(\eta - \mu)^2/(2\sigma^2)}.$$ The solution for $k = 2$ is similar, but more tedious; therefore I have left it as an exercise. The resulting computation is then used to obtain the variance.