Нека $X$ и $Y$ са независими стандартно нормално разпределени случайни величини. Да се пресметне $\text{E}(X|X>Y)$
Решение. Да означим с $F$ и $f$ функциите съответно на разпределение и на плътност на стандартното нормално разпределение (съответно на величините $X$ и $Y$). Нека $Z=X|_{X>Y}$. Така търсим $\text{E}(Z)$. За целта да пресметнем функцията на разпределение $F_Z$ и функцията на плътност $f_Z$. $F_Z(z)=\text{P}(Z<z)=\operatorname{P}(X<z|X>Y)=\frac{\text{P}(X<z \bigcap X>Y)}{P(X>Y)}$. Заради симетрия $\text{P}(X>Y)=\text{P}(Y>X)\Rightarrow \text{P}(X>Y)=\frac{1}{2}$. Ще пресметнем
$\text{P}(X<z \bigcap X>Y)$ чрез формулата за пълна вероятност. Зaбелязваме, че
$$\text{P}(X<z \bigcap X>Y|X=s)=\begin{cases} 0, & s>z \\ \text{P}(Y<s), & s\le z \end{cases}$$ Тогава
$$\text{P}(X<z \bigcap X>Y)=\int_{-\infty}^{\infty}\text{P}(X<z \bigcap X>Y|X=s)f(s)\mathrm{d}s=\int_{-\infty}^{z}F(s)f(s)\mathrm{d}s$$ Така $$F_Z(z)=2\int_{-\infty}^{z}F(s)f(s)\mathrm{d}s$$
За функцията на плътност получаваме $f_Z(z)=2F(z)f(z)$.
Така трябва да пресметнем $$I=\int_{-\infty}^{\infty}zf_Z(z)\mathrm{d}z=2\int_{-\infty}^{\infty}zF(z)f(z)\mathrm{d}z=2\int_{-\infty}^{\infty}zf(z)\int_{-\infty}^{z}f(s)\mathrm{d}s\mathrm{d}z=\\=2\int_{-\infty}^{\infty}\int_{-\infty}^{z}zf(z)f(s)\mathrm{d}s\mathrm{d}z$$
Сега сменяме реда на интегриране $$I/2=\int_{-\infty}^{\infty}\int_{-\infty}^{z}zf(z)f(s)\mathrm{d}s\mathrm{d}z=\int_{-\infty}^{\infty}f(s)\int_{s}^{\infty}zf(z)\mathrm{d}z\mathrm{d}s$$
За вътрешния интеграл намираме
$$\int_{s}^{\infty}zf(z)\mathrm{d}z=\frac{1}{\sqrt{2\pi}}\int_{s}^{\infty}zе^{-\frac{z^2}{2}}\mathrm{d}z=\frac{1}{\sqrt{2\pi}}\int_{s}^{\infty}е^{-\frac{z^2}{2}}\mathrm{d}\frac{z^2}{2}=\frac{1}{\sqrt{2\pi}}\left(-e^{-\frac{z^2}{2}}\right)\Bigg|_{z=s}^{\infty}=f(s)$$
Тогава $$I=2\int_{-\infty}^{\infty}f(s)^2\mathrm{d}s=\frac{1}{\pi}\int_{-\infty}^{\infty}e^{-z^2}\mathrm{d}z=\frac{1}{\sqrt{\pi}}$$

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