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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.)

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

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

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Marcus
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  • we cannot exchange the order of expectation and minimum function, I still dont know the answer, I'll try it, if I get something, I'll share to you – Rizky Reza Fujisaki Nov 04 '16 at 05:40

2 Answers2

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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

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

  • I like your solution far better than mine. – heropup Nov 04 '16 at 05:47
  • There is a confusion between the random variable $x$ and $\eta$. It will be better if you could have taken different variable then $\eta$. – Marcus Nov 04 '16 at 08:39
  • @imtiyazkhan What do you mean by a confusion between $x$ and $\eta$? –  Nov 04 '16 at 08:50
  • @d.k.o As you know I took $\eta$ as a constant. But you included $\eta$ in your variable $z_{\eta}$ as another variable. It makes me confuse which $\eta$ is a variable and which $\eta$ is constant – Marcus Nov 04 '16 at 08:56
  • @imtiyazkhan $z_{\eta}$ is also a constant. $\eta$ in its definition is the same $\eta$ you used in the question... –  Nov 04 '16 at 09:04
  • @d.k.o. Thanks. One more thing is that make me uncomfortable when I see $X1, X^21$. Why you included those $1$ everywhere. Please also give me the reference from you took or inspired for this answer. – Marcus Nov 04 '16 at 09:26
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    @imtiyazkhan $1_A$ is the indicator of set $A$. So $X1{X\le \eta}=X$ on the set ${X\le \eta}$ and is $0$ otherwise. Also $$ X\wedge \eta=(X\wedge \eta)1{X\le \eta}+(X\wedge \eta)1{X> \eta} \ =X\times 1{X\le \eta}+\eta\times 1{X> \eta}. $$ –  Nov 04 '16 at 15:39
  • @d.k.o. Thanks ) It helps. But what about reference?? – Marcus Nov 05 '16 at 14:27
  • @imtiyazkhan Actually, it is a standard trick. I'm not sure what type of reference you are asking about? –  Nov 05 '16 at 18:09
  • @d.k.o. Thanks. Your answer helped me alot. Now first line is understandable to me. But how do you get $m\Phi(z_{\eta})-\sigma\phi(z_{\eta})+\eta(1-\Phi(z_{\eta}))$ from $E[X1{X\le \eta}]+\eta \mathbb{E}[1{X>\eta}] $ – Marcus Nov 08 '16 at 11:27
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    $$\mathbb{E}[1{X>\eta}]=\mathbb{P}{X>\eta}=1-\Phi(z_\eta),$$

    $$\mathbb{E}[X1{X\le\eta}]=\mathbb{E}[X\mid X\le\eta]\times \mathbb{P}{X\le \eta}$$

    For $\mathbb{E}[X\mid X\le\eta]$ you may look here:https://en.wikipedia.org/wiki/Truncated_normal_distribution.

    Otherwise, you may calculate $\mathbb{E}[X1{X\le\eta}]$ directly (as heropup showed).

     –  Nov 09 '16 at 04:33
  • Thanks alot )). Please help me to answer my this question as well http://math.stackexchange.com/questions/2007971/mean-and-variance-of-random-variable-having-min-function – Marcus Nov 10 '16 at 13:13
  • Hi, Can you help me how you have calculated \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} because when I am trying to find in the simailar fashion I am not getting the desired result. – Marcus Nov 23 '16 at 10:21
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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.

heropup
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  • Hi, Please help me to answer my this question as well http://math.stackexchange.com/questions/2007971/variance-of-random-variable-having-min-function?noredirect=1#comment4123131_2007971 – Marcus Nov 11 '16 at 05:30