Computing E [ X ∣ X > Y ] , E[X \mid X > Y], E [ X ∣ X > Y ] , Where X , Y ∼ N ( 0 , 1 ) X, Y \sim \mathcal{N}(0, 1) X , Y ∼ N ( 0 , 1 )

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Introduction

In probability theory, conditional expectation is a fundamental concept that deals with the expected value of a random variable given that another random variable has taken on a specific value. In this article, we will explore the computation of the conditional expectation of $X$ given that $X$ is greater than $Y$, where both $X$ and $Y$ are normally distributed with a mean of 0 and a variance of 1.

Background

To begin with, let's recall the definition of conditional expectation. Given two random variables $X$ and $Y$, the conditional expectation of $X$ given $Y$ is denoted by $E[X \mid Y]$ and is defined as the expected value of $X$ when $Y$ is known. In other words, it is the average value of $X$ that we would expect to observe if we knew the value of $Y$.

The Case of Fixed $Y$

When $Y$ is fixed to a constant $c$, the conditional expectation of $X$ given $X > Y$ can be computed using the following formula:

$$E[X \mid X > Y = c] = \frac{\int_{c}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx}{\int_{c}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx}$$

This formula can be simplified by recognizing that the denominator is the cumulative distribution function (CDF) of the standard normal distribution, which is given by:

$$\Phi(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}} dt$$

Using this fact, we can rewrite the numerator as:

$$\int_{c}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx = \int_{c}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx \cdot x$$

Substituting this expression into the original formula, we get:

$$E[X \mid X > Y = c] = \frac{\int_{c}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx \cdot x}{\int_{c}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx}$$

Simplifying further, we get:

$$E[X \mid X > Y = c] = c + \sqrt{\frac{2}{\pi}}$$

The General Case

In the general case where $Y$ is not fixed to a constant, we need to compute the conditional expectation of $X$ given $X > Y$ using the following formula:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{P(X > Y)}{P(X > Y)} dF_Y(x)$$

where $P(X > Y)$ is the probability that $X$ is greater than $Y$, and $dF_Y(x)$ is the probability density function (PDF) of $Y$.

Computing $P(X > Y)$

To compute $P(X > Y)$, we need to integrate the joint PDF of $X$ and $Y$ over the region where $X > Y$. Since $X$ and $Y$ are independent and identically distributed (i.i.d.) standard normal random variables, their joint PDF is given by:

$$f_{X,Y}(x,y) = \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}}$$

Integrating this PDF over the region where $X > Y$, we get:

$$P(X > Y) = \int_{-\infty}^{\infty} \int_{y}^{\infty} \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}} dx dy$$

Evaluating this integral, we get:

$$P(X > Y) = \frac{1}{2}$$

Computing $E[X \mid X > Y]$

Now that we have computed $P(X > Y)$, we can compute the conditional expectation of $X$ given $X > Y$ using the following formula:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{P(X > Y)}{P(X > Y)} dF_Y(x)$$

Substituting the value of $P(X > Y)$, we get:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{1/2}{1/2} dF_Y(x)$$

Simplifying further, we get:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x dF_Y(x)$$

Evaluating the Integral

To evaluate the integral, we need to find the PDF of $Y$. Since $Y$ is a standard normal random variable, its PDF is given by:

$$f_Y(y) = \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}}$$

Substituting this expression into the integral, we get:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} dy$$

Evaluating this integral, we get:

$$E[X \mid X > Y] = \sqrt{\frac{2}{\pi}}$$

Conclusion

In this article, we have computed the conditional expectation of $X$ given $X > Y$, where $X$ and $Y$ are i.i.d. standard normal random variables. We have shown that when $Y$ is fixed to a constant $c$, the conditional expectation of $X$ given $X > Y$ is given by $c + \sqrt{\frac{2}{\pi}}$. In the general case where $Y$ is not fixed, we have shown that the conditional expectation of $X$ given $X > Y$ is given by $\sqrt{\frac{2}{\pi}}$.

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Introduction

In our previous article, we explored the computation of the conditional expectation of $X$ given that $X$ is greater than $Y$, where both $X$ and $Y$ are normally distributed with a mean of 0 and a variance of 1. In this article, we will answer some frequently asked questions related to this topic.

Q1: What is the relationship between the conditional expectation of $X$ given $X > Y$ and the probability that $X$ is greater than $Y$?

A1: The conditional expectation of $X$ given $X > Y$ is related to the probability that $X$ is greater than $Y$ through the formula:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{P(X > Y)}{P(X > Y)} dF_Y(x)$$

where $P(X > Y)$ is the probability that $X$ is greater than $Y$, and $dF_Y(x)$ is the probability density function (PDF) of $Y$.

Q2: How do we compute the probability that $X$ is greater than $Y$?

A2: To compute the probability that $X$ is greater than $Y$, we need to integrate the joint PDF of $X$ and $Y$ over the region where $X > Y$. Since $X$ and $Y$ are independent and identically distributed (i.i.d.) standard normal random variables, their joint PDF is given by:

$$f_{X,Y}(x,y) = \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}}$$

Integrating this PDF over the region where $X > Y$, we get:

$$P(X > Y) = \int_{-\infty}^{\infty} \int_{y}^{\infty} \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}} dx dy$$

Evaluating this integral, we get:

$$P(X > Y) = \frac{1}{2}$$

Q3: What is the relationship between the conditional expectation of $X$ given $X > Y$ and the mean of $X$?

A3: The conditional expectation of $X$ given $X > Y$ is related to the mean of $X$ through the formula:

$$E[X \mid X > Y] = \sqrt{\frac{2}{\pi}}$$

This means that the conditional expectation of $X$ given $X > Y$ is a constant that is independent of the mean of $X$.

Q4: Can we use the same formula to compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are not i.i.d. standard normal random variables?

A4: No, we cannot use the same formula to compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are not i.i.d. standard normal random variables. The formula we used in this article is specific to the case where $X$ and $Y$ are i.i.d. standard normal random variables.

Q5: How do we compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are not i.i.d. standard normal random variables?

A5: To compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are not i.i.d. standard normal random variables, we need to use a different approach. We can use the formula:

$$E[X \mid X > Y] = \int_{-\infty}^{\infty} x \frac{P(X > Y)}{P(X > Y)} dF_Y(x)$$

where $P(X > Y)$ is the probability that $X$ is greater than $Y$, and $dF_Y(x)$ is the PDF of $Y$. We need to compute the probability that $X$ is greater than $Y$ and the PDF of $Y$ using the specific distributions of $X$ and $Y$.

Conclusion

In this article, we have answered some frequently asked questions related to the computation of the conditional expectation of $X$ given that $X$ is greater than $Y$, where both $X$ and $Y$ are normally distributed with a mean of 0 and a variance of 1. We have shown that the conditional expectation of $X$ given $X > Y$ is related to the probability that $X$ is greater than $Y$ and the mean of $X$. We have also discussed how to compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are not i.i.d. standard normal random variables.