A Special Case That Slutsky's Theorem Can Be Conversed

Introduction

Slutsky's theorem is a fundamental concept in probability theory that deals with the convergence of random variables. It states that if a sequence of random variables $X_n$ converges in probability to a random variable $X$, and another sequence of random variables $Y_n$ converges in probability to a constant $c$, then the sequence $X_n + Y_n$ converges in distribution to $X + c$. However, there are certain special cases where Slutsky's theorem can be converted to provide more insight into the behavior of the random variables. In this article, we will explore one such special case.

The Problem

We are given two sequences of random variables $X_n$ and $Y_n$ such that $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$. We are also given that $X_n + Y_n \xrightarrow{d} N(0,1)$. The question is whether it is true that $Y_n \xrightarrow{d} 0$.

Slutsky's Theorem

Before we dive into the special case, let's recall Slutsky's theorem. If $X_n \xrightarrow{p} X$ and $Y_n \xrightarrow{p} c$, then $X_n + Y_n \xrightarrow{d} X + c$. Here, $\xrightarrow{p}$ denotes convergence in probability, and $\xrightarrow{d}$ denotes convergence in distribution.

The Special Case

In this special case, we have $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$. We are also given that $X_n + Y_n \xrightarrow{d} N(0,1)$. To determine whether $Y_n \xrightarrow{d} 0$, we can use the following approach:

Step 1: Use the Continuous Mapping Theorem

The continuous mapping theorem states that if $X_n \xrightarrow{d} X$ and $g$ is a continuous function, then $g(X_n) \xrightarrow{d} g(X)$. In this case, we can define a function $g(x) = x + y$, where $y$ is a constant. Then, we have $g(X_n) = X_n + Y_n$, which converges in distribution to $N(0,1)$.

Step 2: Use the Delta Method

The delta method states that if $X_n \xrightarrow{d} X$ and $g$ is a differentiable function, then $g(X_n) \xrightarrow{d} g(X) + g'(X) \cdot (X_n - X)$. In this case, we can define a function $g(x) = x + y$, where $y$ is a constant. Then, we have $g(X_n) = X_n + Y_n$, which converges in distribution to $N(0,1)$.

Step 3: Use the Slutsky's Theorem

Slutsky's theorem states that if $X_n \xrightarrow{p} X$ and $Y_n \xrightarrow{p} c$, then $X_n + Y_n \xrightarrow{d} X + c$. In this case, we have $X_n \xrightarrow{d}(0,1)$ and $Y_n \geq 0$. We are also given that $X_n + Y_n \xrightarrow{d} N(0,1)$. Therefore, we can conclude that $Y_n \xrightarrow{d} 0$.

Conclusion

In this article, we explored a special case where Slutsky's theorem can be converted to provide more insight into the behavior of the random variables. We showed that if $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$, and $X_n + Y_n \xrightarrow{d} N(0,1)$, then $Y_n \xrightarrow{d} 0$. This result has important implications in probability theory and statistics.

References

• Billingsley, P. (1995). Probability and Measure. John Wiley & Sons.
• Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury Press.
• Feller, W. (1971). An Introduction to Probability Theory and Its Applications. John Wiley & Sons.

Further Reading

• Continuous Mapping Theorem
• Delta Method
• Slutsky's Theorem

Introduction

In our previous article, we explored a special case where Slutsky's theorem can be converted to provide more insight into the behavior of the random variables. We showed that if $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$, and $X_n + Y_n \xrightarrow{d} N(0,1)$, then $Y_n \xrightarrow{d} 0$. In this article, we will answer some frequently asked questions related to this special case.

Q: What is Slutsky's theorem?

A: Slutsky's theorem is a fundamental concept in probability theory that deals with the convergence of random variables. It states that if a sequence of random variables $X_n$ converges in probability to a random variable $X$, and another sequence of random variables $Y_n$ converges in probability to a constant $c$, then the sequence $X_n + Y_n$ converges in distribution to $X + c$.

Q: What is the continuous mapping theorem?

A: The continuous mapping theorem states that if $X_n \xrightarrow{d} X$ and $g$ is a continuous function, then $g(X_n) \xrightarrow{d} g(X)$. This theorem is often used to prove the convergence of random variables under certain conditions.

Q: What is the delta method?

A: The delta method states that if $X_n \xrightarrow{d} X$ and $g$ is a differentiable function, then $g(X_n) \xrightarrow{d} g(X) + g'(X) \cdot (X_n - X)$. This theorem is often used to prove the convergence of random variables under certain conditions.

Q: Can we use Slutsky's theorem to prove the convergence of $Y_n$?

A: Yes, we can use Slutsky's theorem to prove the convergence of $Y_n$. Since $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$, and $X_n + Y_n \xrightarrow{d} N(0,1)$, we can conclude that $Y_n \xrightarrow{d} 0$.

Q: What are the implications of this result?

A: This result has important implications in probability theory and statistics. It shows that if a sequence of random variables $X_n$ converges in distribution to a normal distribution, and another sequence of random variables $Y_n$ is non-negative, then the sequence $X_n + Y_n$ converges in distribution to a normal distribution.

Q: Can we generalize this result to other distributions?

A: Yes, we can generalize this result to other distributions. However, the proof of the result would require additional assumptions and conditions.

Q: What are some common applications of this result?

A: This result has many applications in probability theory and statistics. Some common applications include:

• Hypothesis testing: This result is often used in hypothesis testing to determine whether a sequence of random variables converges to a certain distribution.
• Confidence intervals: result is often used in confidence intervals to determine the convergence of a sequence of random variables.
• Regression analysis: This result is often used in regression analysis to determine the convergence of a sequence of random variables.

Conclusion

In this article, we answered some frequently asked questions related to the special case where Slutsky's theorem can be converted. We showed that if $X_n \xrightarrow{d} N(0,1)$ and $Y_n \geq 0$, and $X_n + Y_n \xrightarrow{d} N(0,1)$, then $Y_n \xrightarrow{d} 0$. This result has important implications in probability theory and statistics.

References

• Billingsley, P. (1995). Probability and Measure. John Wiley & Sons.
• Casella, G., & Berger, R. L. (2002). Statistical Inference. Duxbury Press.
• Feller, W. (1971). An Introduction to Probability Theory and Its Applications. John Wiley & Sons.

Further Reading

• Continuous Mapping Theorem
• Delta Method
• Slutsky's Theorem