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 some additional information. 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 mean 0 and variance 1. This problem is a classic example of conditional expectation and has important applications in statistics and probability theory.

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 $X$ given the value of $Y$. In other words, it is the average value of $X$ that we would expect to observe given the value of $Y$.

The Case When $Y = c$

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

$$E[X \mid X > c] = \frac{\int_{c}^{\infty} x f(x) dx}{\int_{c}^{\infty} f(x) dx}$$

where $f(x)$ is the probability density function of $X$. In the case of a normal distribution with mean 0 and variance 1, the probability density function is given by:

$$f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$$

Substituting this into the formula above, we get:

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

The General Case

In the general case where $Y$ is not a constant, but rather a random variable with its own distribution, the computation of the conditional expectation of $X$ given $X > Y$ is more involved. We can use the following formula:

$$E[X \mid X > Y] = \frac{\int_{-\infty}^{\infty} x f_{X \mid Y}(x \mid y) f_Y(y) dy dx}{\int_{-\infty}^{\infty} f_{X \mid Y}(x \mid y) f_Y(y) dy dx}$$

where $f_{X \mid Y}(x \mid y)$ is the conditional probability density function of $X$ given $Y$, and $f_Y(y)$ is the probability density function of $Y$.

Computation of the Conditional Probability Density Function

To compute the conditional probability density function of $X$ given $Y$, we can use the following formula:

$$f_{X \mid Y}(x \mid y) = \frac{f(x, y)}{f_Y(y)}$$

where $f(x, y)$ is the joint probability density function of $X$ and $Y$. In the case of two independent normal random variables with mean 0 and variance 1, the joint probability density function is given by:

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

Substituting this into the formula above, we get:

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

Computation of the Conditional Expectation

Now that we have the conditional probability density function of $X$ given $Y$, we can compute the conditional expectation of $X$ given $X > Y$ using the following formula:

$$E[X \mid X > Y] = \frac{\int_{-\infty}^{\infty} x f_{X \mid Y}(x \mid y) f_Y(y) dy dx}{\int_{-\infty}^{\infty} f_{X \mid Y}(x \mid y) f_Y(y) dy dx}$$

Substituting the expression for $f_{X \mid Y}(x \mid y)$ above, we get:

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

Simplification of the Expression

To simplify the expression above, we can use the following substitution:

$$u = \frac{x}{\sqrt{2}}$$

$$v = \frac{y}{\sqrt{2}}$$

This gives us:

$$E[X \mid X > Y] = \frac{\int_{-\infty}^{\infty} \sqrt{2} u \frac{1}{\sqrt{\pi}} e^{-u^2} \frac{1}{\sqrt{\pi}} e^{-v^2} dv du}{\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi}} e^{-u^2} \frac{1}{\sqrt{\pi}} e^{-v^2} dv du}$$

Final Expression

After simplifying the expression above, we get:

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

This is the final expression for the conditional expectation of $X$ given $X > Y$, where $X$ and $Y$ are normally distributed with mean 0 and variance 1.

Conclusion

In this article, we have computed the conditional expectation of $X$ given $X > Y$, where $X$ and $Y$ are normally distributed with mean 0 and variance 1. We have used the formula for conditional expectation and the properties of the normal distribution to derive the final expression. The result shows that the conditional expectation of $X$ given $X > Y$ is equal to $\sqrt{2}$.

References

• [1] Ross, S. M. (2010). Introduction to Probability Models. Academic Press.
• [2] Papoulis, A. (2002). Probability, Random Variables, and Stochastic Processes. McGraw-Hill.
• [3] Johnson, N. L., & Kotz, S. (1970). Distributions in Statistics: Continuous Univariate Distributions. Wiley.
  

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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 mean 0 and variance 1. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the significance of the conditional expectation in probability theory?

A: The conditional expectation is a fundamental concept in probability theory that deals with the expected value of a random variable given some additional information. It is used to make predictions and decisions in various fields such as statistics, engineering, and economics.

Q: How do we compute the conditional expectation of $X$ given $X > Y$ when $X$ and $Y$ are independent normal random variables?

A: We can use the formula for conditional expectation and the properties of the normal distribution to derive the final expression. The result shows that the conditional expectation of $X$ given $X > Y$ is equal to $\sqrt{2}$.

Q: What is the role of the joint probability density function in computing the conditional expectation?

A: The joint probability density function of $X$ and $Y$ is used to compute the conditional probability density function of $X$ given $Y$. This is then used to compute the conditional expectation of $X$ given $X > Y$.

Q: Can we generalize the result to other types of distributions?

A: Yes, the result can be generalized to other types of distributions. However, the computation of the conditional expectation may be more involved and may require the use of different techniques and formulas.

Q: What are some applications of the conditional expectation in real-world problems?

A: The conditional expectation has many applications in real-world problems such as:

• Predicting stock prices based on historical data
• Making decisions in engineering and economics
• Analyzing data in medicine and social sciences

Q: How do we handle the case when $Y$ is not a constant, but rather a random variable with its own distribution?

A: We can use the formula for conditional expectation and the properties of the normal distribution to derive the final expression. The result shows that the conditional expectation of $X$ given $X > Y$ is equal to $\sqrt{2}$.

Q: What are some common mistakes to avoid when computing the conditional expectation?

A: Some common mistakes to avoid when computing the conditional expectation include:

• Failing to account for the joint distribution of $X$ and $Y$
• Using the wrong formula or technique
• Not considering the properties of the normal distribution

Q: How do we verify the result and ensure that it is correct?

A: We can verify the result by checking the following:

• The formula for conditional expectation is correct
• The properties of the normal distribution are correctly applied
• The result is consistent with the expected value of $X$ given $X > Y$

Conclusion

In this article, we have answered some frequently asked questions related to the computation of the conditional expectation of $X$ given $X > Y$, where $X$ and $Y$ are normally distributed with mean 0 and variance 1. We hope that this article has provided a clear understanding of the concept and its applications.

References

• [1] Ross, S. M. (2010). Introduction to Probability Models. Academic Press.
• [2] Papoulis, A. (2002). Probability, Random Variables, and Stochastic Processes. McGraw-Hill.
• [3] Johnson, N. L., & Kotz, S. (1970). Distributions in Statistics: Continuous Univariate Distributions. Wiley.