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Let $Y_1<Y_2$ be order statistics from a random sample of size $2$ from a normal distribution, $\mathcal{N}(\mu,\sigma^2)$, where $\sigma^2$ is known. Show that $P(Y_1<\mu<Y_2)=\frac12$ and find $E(Y_1-Y_2)$.

I am not exactly sure how to solve the question above. Any help would be appreciated. Thanks.

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$P(Y_1\lt\mu\lt Y_2)=0$, not $\frac12$. – Did Nov 13 '11 at 10:31
@Didier In other words, that's a typo. The "1" and the "2" got switched. – Michael Hardy Nov 13 '11 at 13:26
I've gone ahead and edited the question to fix the typo. – Michael Hardy Nov 13 '11 at 13:27
icobes: keep in mind that there is a whole site,, dedicated to statistics! – Mariano Suárez-Alvarez Nov 13 '11 at 14:21
@Michael: No kidding? I never would have guessed. – Did Nov 13 '11 at 14:22
up vote 3 down vote accepted

Let $X_1,X_2$ be the i.i.d. sample; then $Y_2 =\max\{X_1,X_2\}$ and $Y_1=\min\{X_1,X_2\}$ (I'm regarding "${<}$" as a typo in the question, in view of the result to be proved).

Then either $Y_1<Y_2<\mu$ or $Y_1<\mu<Y_2$ or $\mu<Y_1<Y_2$. (I'm discounting the event of probability $0$ that two or more of these are equal.)

The first happens if and only if both $X_1$ and $X_2$ are less than $\mu$; the second if and only if one (either one) is less than $\mu$ and the other greater; the third if and only if both are greater than $\mu$.

The probability that $X_1>\mu$ is $1/2$; similarly for $X_2$.

So the event $Y_1<\mu<Y_2$ is the event of exactly one success in two independent trials, with probability $1/2$ of success on each trial. Therefore its probability is $1/2$.

Now notice that $E(Y_2-Y_1) = E(|X_2-X_1|)$, and $X_1-X_2 \sim \mathcal{N}(\mu-\mu,\sigma^2+\sigma^2)=\mathcal{N}(0,2\sigma^2)$. So $E(|X_2-X_1|)= \sqrt{2}\sigma E\left(\dfrac{|X_2-X_1|}{\sqrt{2}\sigma}\right)$ and $Z=\dfrac{X_2-X_1}{\sqrt{2}\sigma}\sim\mathcal{N}(0,1)$. So we want $\sqrt{2}\sigma E(|Z|)$.

So $$ \begin{align} E(|Z|) & = \int_{-\infty}^\infty |z| \varphi(z)\;dz = 2\int_0^\infty z \varphi(z)\;dz = 2\int_0^\infty z \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \; dz \\ \\ & = \sqrt{\frac{2}{\pi}} \int_0^\infty ze^{-z^2/2} \; dz = \sqrt{\frac{2}{\pi}} \int_0^\infty e^{-u} \; du = \sqrt{\frac{2}{\pi}}. \end{align} $$

Multiplying that by $\sqrt{2}\;\sigma$, we get $\dfrac{2\sigma}{\sqrt{\pi}}$.

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The hypothesis should be $Y_1 < Y_2$.

Now the mutual density is given by : $f_{Y_1,Y_2}(y_1,y_2) = 2 f_{X}(y_1) f_{X}(y_2)$ for any $y_1,y_2$ such that $y_1 < y_2$, and $f_{Y_1,Y_2}(y_1,y_2) = 0$ otherwise. $f_X$ is the density of $\mathcal{N}(\mu,\sigma)$.

So : $P(Y_1 < \mu < Y_2) = \int_{D} f_{Y_1,Y_2}(y_1,y_2)dy_1dy_2$ on the domain $D=\left\{y_1 < \mu < y_2 \right\}$. If I take a pen and paper, draw this domain, I find : $f_{Y_1,Y_2}(y_1,y_2) = 1 - \int_{-\infty}^{\mu}\int_\mu^{+\infty}2 f_{X}(y_1) f_{X}(y_2)dy_1dy_2 = 1-(2F_{X}(\mu))(1-F_{X}(\mu))=\frac12$ since $F_{X}(\mu) = F_{X}(\mu)=\frac12$.

E(Y_1-Y_2) = E(Y_1) - E(Y_2) = 0

edit Oups, I wrote this too fast! $Y_1$ and $Y_2$ don't have the same expectation.

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Mmhhh the solution to the second part of the question is very wrong. – Did Nov 13 '11 at 11:29
I'm surprised at how complicated you make this. The event in question is simply the event that the number of successes in two independent trials is $1$, i.e. exactly one of $X_1,X_2$ is less than $\mu$. And the probability of success on each trial is $1/2$. – Michael Hardy Nov 13 '11 at 13:02

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