daniko883y

2022-09-08

What is a distribution of X given $X-Y>0$

given two normally distributed variables $X\sim \mathcal{N}({\mu}_{X},{\sigma}_{X})$ and $Y\sim \mathcal{N}({\mu}_{Y},{\sigma}_{Y})$ that are independent ${\rho}_{XY}=0$, what is a distribution of X given $X>Y$.

I have run several simulations in Matlab and $X|X>Y$ looks suspiciously normally distributed, but what are its mean and standard deviation?

given two normally distributed variables $X\sim \mathcal{N}({\mu}_{X},{\sigma}_{X})$ and $Y\sim \mathcal{N}({\mu}_{Y},{\sigma}_{Y})$ that are independent ${\rho}_{XY}=0$, what is a distribution of X given $X>Y$.

I have run several simulations in Matlab and $X|X>Y$ looks suspiciously normally distributed, but what are its mean and standard deviation?

Alvin Preston

Beginner2022-09-09Added 9 answers

Step 1

Instead of thinking through the inequality $X>Y$, see it as $Z=X-Y>0$. Note that given that X and Y are both normal, Z is normal with mean ${\mu}_{x}-{\mu}_{Y}$ and variance ${\sigma}_{X}^{2}+{\sigma}_{Y}^{2}$ (assuming independence). Thus, you can calculate

$$\begin{array}{rl}P\{X\le x|X>Y\}& =\frac{P\{X\le x,X>Y\}}{P\{X>Y\}}\\ & =\frac{P\{Y<X\le x\}}{P\{Z>0\}}\end{array}$$

I assume you can work out $P\{Z>0\}$ with ease, for it is the distribution function of a normal r.v.. The hard part is $P\{Y<X\le x\}$. You can draw the area in ${\mathbb{R}}^{2}$ to see how to integrate it. I believe it is as follows (I changed x for z so there is no confusion in the integrands and limits):

$$\begin{array}{rl}P\{Y<X\le z\}& ={\int}_{-\mathrm{\infty}}^{z}{\int}_{-\mathrm{\infty}}^{x}f(x,y)dydx\\ & ={\int}_{-\mathrm{\infty}}^{z}{\int}_{-\mathrm{\infty}}^{x}f(x)f(y)dydx\phantom{\rule{1em}{0ex}}(\text{Independence})\\ & ={\int}_{-\mathrm{\infty}}^{z}f(x){\int}_{-\mathrm{\infty}}^{x}f(y)dydx\\ & ={\int}_{-\mathrm{\infty}}^{z}f(x){F}_{Y}(x)dx\end{array}$$

Step 2

However, we could try to solve the simplest case, which is that where X and Y are standard normal r.v.. In such case, we have that

$$\begin{array}{rl}P\{Y<X\le z\}& ={\int}_{-\mathrm{\infty}}^{z}f(x){F}_{Y}(x)dx\\ & ={\int}_{-\mathrm{\infty}}^{z}\varphi (x)\mathrm{\Phi}(x)dx\\ & ={\int}_{0}^{\mathrm{\Phi}(z)}udu=0.5\mathrm{\Phi}(z{)}^{2}\end{array}$$

where we used the variable change $u=\mathrm{\Phi}(x)$, and thus $du=\varphi (x)dx$, where $\mathrm{\Phi}$ is the distribution function of a standard normal and $\varphi $ its density. Moreover, now Z has mean 0 so $P(Z>0)=0.5$. Thus,

$$P\{X\le x|X>Y\}=0.5\mathrm{\Phi}(x{)}^{2}/0.5=\mathrm{\Phi}(x{)}^{2}.$$

Then, $f(x|X>Y)=2\mathrm{\Phi}(x)\varphi (x)$. The image below shows a plot of the standard normal density $\varphi (x)$ and $f(x|X>Y)$ as given before. You can see they look very similar, as you guessed through simulation. However, the r.v. $X|X>Y$ is not normal even under the simplest of assumptions.

Instead of thinking through the inequality $X>Y$, see it as $Z=X-Y>0$. Note that given that X and Y are both normal, Z is normal with mean ${\mu}_{x}-{\mu}_{Y}$ and variance ${\sigma}_{X}^{2}+{\sigma}_{Y}^{2}$ (assuming independence). Thus, you can calculate

$$\begin{array}{rl}P\{X\le x|X>Y\}& =\frac{P\{X\le x,X>Y\}}{P\{X>Y\}}\\ & =\frac{P\{Y<X\le x\}}{P\{Z>0\}}\end{array}$$

I assume you can work out $P\{Z>0\}$ with ease, for it is the distribution function of a normal r.v.. The hard part is $P\{Y<X\le x\}$. You can draw the area in ${\mathbb{R}}^{2}$ to see how to integrate it. I believe it is as follows (I changed x for z so there is no confusion in the integrands and limits):

$$\begin{array}{rl}P\{Y<X\le z\}& ={\int}_{-\mathrm{\infty}}^{z}{\int}_{-\mathrm{\infty}}^{x}f(x,y)dydx\\ & ={\int}_{-\mathrm{\infty}}^{z}{\int}_{-\mathrm{\infty}}^{x}f(x)f(y)dydx\phantom{\rule{1em}{0ex}}(\text{Independence})\\ & ={\int}_{-\mathrm{\infty}}^{z}f(x){\int}_{-\mathrm{\infty}}^{x}f(y)dydx\\ & ={\int}_{-\mathrm{\infty}}^{z}f(x){F}_{Y}(x)dx\end{array}$$

Step 2

However, we could try to solve the simplest case, which is that where X and Y are standard normal r.v.. In such case, we have that

$$\begin{array}{rl}P\{Y<X\le z\}& ={\int}_{-\mathrm{\infty}}^{z}f(x){F}_{Y}(x)dx\\ & ={\int}_{-\mathrm{\infty}}^{z}\varphi (x)\mathrm{\Phi}(x)dx\\ & ={\int}_{0}^{\mathrm{\Phi}(z)}udu=0.5\mathrm{\Phi}(z{)}^{2}\end{array}$$

where we used the variable change $u=\mathrm{\Phi}(x)$, and thus $du=\varphi (x)dx$, where $\mathrm{\Phi}$ is the distribution function of a standard normal and $\varphi $ its density. Moreover, now Z has mean 0 so $P(Z>0)=0.5$. Thus,

$$P\{X\le x|X>Y\}=0.5\mathrm{\Phi}(x{)}^{2}/0.5=\mathrm{\Phi}(x{)}^{2}.$$

Then, $f(x|X>Y)=2\mathrm{\Phi}(x)\varphi (x)$. The image below shows a plot of the standard normal density $\varphi (x)$ and $f(x|X>Y)$ as given before. You can see they look very similar, as you guessed through simulation. However, the r.v. $X|X>Y$ is not normal even under the simplest of assumptions.

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