Expected Value Of Highest Variable So Far

Introduction

In probability theory, the concept of expected value is a fundamental idea that helps us understand the average behavior of random variables. The expected value of a random variable is a measure of its central tendency, and it plays a crucial role in many areas of statistics and probability. In this article, we will explore the expected value of the highest variable so far in a sequence of random variables.

Background

Let $X_1,X_2,\ldots$ be iid (independent and identically distributed) random variables uniformly distributed over $[0,1]$. This means that each $X_i$ is a random variable that takes on a value between 0 and 1, and the probability distribution of each $X_i$ is the same. We are interested in finding the expected value of the highest variable so far in this sequence.

Definition of N

Let $N$ be the first position in the sequence where the number at that position is not the highest number seen so far. In other words, $N$ is the position where the sequence starts to decrease. For example, if the sequence is $X_1 = 0.5, X_2 = 0.7, X_3 = 0.3, X_4 = 0.9$, then $N = 3$ because the sequence starts to decrease at the third position.

Expected Value of N

To find the expected value of $N$, we need to find the probability distribution of $N$. Let $P(N = k)$ be the probability that $N = k$. We can write this as:

$$P(N = k) = P(X_1 \leq X_2 \leq \ldots \leq X_k < X_{k+1})$$

Using the fact that the $X_i$ are iid, we can rewrite this as:

$$P(N = k) = P(X_1 \leq X_2 \leq \ldots \leq X_k) \cdot P(X_{k+1} > X_k)$$

The first term on the right-hand side is the probability that the first $k$ variables are in increasing order, which is equal to $\frac{1}{k!}$. The second term is the probability that the $(k+1)$th variable is greater than the $k$th variable, which is equal to $\frac{1}{2}$.

Calculating the Expected Value of N

Using the probability distribution of $N$, we can calculate the expected value of $N$ as follows:

$$E[N] = \sum_{k=1}^{\infty} k \cdot P(N = k)$$

Substituting the expression for $P(N = k)$, we get:

$$E[N] = \sum_{k=1}^{\infty} k \cdot \frac{1}{k!} \cdot \frac{1}{2}$$

Simplifying this expression, we get:

$$E[N] = \frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{(k-1)!}$$

This is a well-known series, and its sum is equal to $e$.

Expected Value of Highest Variable So Far

Now that we have found the expected of $N$, we can find the expected value of the highest variable so far. Let $X_{N-1}$ be the highest variable so far. Then, the expected value of $X_{N-1}$ is equal to:

$$E[X_{N-1}] = E[X_1] + E[X_2] + \ldots + E[X_{N-1}]$$

Using the fact that the $X_i$ are iid, we can rewrite this as:

$$E[X_{N-1}] = (N-1) \cdot E[X_1]$$

Since $E[X_1] = \frac{1}{2}$, we have:

$$E[X_{N-1}] = (N-1) \cdot \frac{1}{2}$$

Substituting the expression for $E[N]$, we get:

$$E[X_{N-1}] = (e-1) \cdot \frac{1}{2}$$

Conclusion

In this article, we have explored the expected value of the highest variable so far in a sequence of random variables. We have found that the expected value of the highest variable so far is equal to $(e-1) \cdot \frac{1}{2}$. This result has important implications for many areas of statistics and probability, and it highlights the importance of understanding the concept of expected value.

References

• [1] Feller, W. (1968). An Introduction to Probability Theory and Its Applications. John Wiley & Sons.
• [2] Grimmett, G. R., & Stirzaker, D. R. (2001). Probability and Random Processes. Oxford University Press.
• [3] Ross, S. M. (2014). Introduction to Probability Models. Academic Press.

Further Reading

• [1] Probability and Statistics: A Course for Graduate Students. By David R. Brillinger.
• [2] Probability Theory: The Logic of Science. By E. T. Jaynes.
• [3] Random Processes: A First Course. By I. I. Gikhman and A. V. Skorokhod.
  Expected Value of Highest Variable So Far: Q&A =====================================================

Introduction

In our previous article, we explored the expected value of the highest variable so far in a sequence of random variables. We found that the expected value of the highest variable so far is equal to $(e-1) \cdot \frac{1}{2}$. In this article, we will answer some frequently asked questions about this topic.

Q: What is the expected value of the highest variable so far?

A: The expected value of the highest variable so far is equal to $(e-1) \cdot \frac{1}{2}$.

Q: How did you derive this result?

A: We derived this result by first finding the probability distribution of the first position in the sequence where the number at that position is not the highest number seen so far. We then used this probability distribution to find the expected value of the highest variable so far.

Q: What is the significance of this result?

A: This result has important implications for many areas of statistics and probability. For example, it can be used to understand the behavior of random variables in various contexts, such as finance, engineering, and computer science.

Q: Can you provide more details about the probability distribution of the first position in the sequence where the number at that position is not the highest number seen so far?

A: The probability distribution of the first position in the sequence where the number at that position is not the highest number seen so far is given by:

$$P(N = k) = P(X_1 \leq X_2 \leq \ldots \leq X_k < X_{k+1})$$

Using the fact that the $X_i$ are iid, we can rewrite this as:

$$P(N = k) = P(X_1 \leq X_2 \leq \ldots \leq X_k) \cdot P(X_{k+1} > X_k)$$

The first term on the right-hand side is the probability that the first $k$ variables are in increasing order, which is equal to $\frac{1}{k!}$. The second term is the probability that the $(k+1)$th variable is greater than the $k$th variable, which is equal to $\frac{1}{2}$.

Q: Can you provide more details about the expected value of the highest variable so far?

A: The expected value of the highest variable so far is given by:

$$E[X_{N-1}] = (N-1) \cdot E[X_1]$$

Since $E[X_1] = \frac{1}{2}$, we have:

$$E[X_{N-1}] = (N-1) \cdot \frac{1}{2}$$

Substituting the expression for $E[N]$, we get:

$$E[X_{N-1}] = (e-1) \cdot \frac{1}{2}$$

Q: What are some applications of this result?

A: This result has many applications in various fields, such as:

• Finance: Understanding the behavior of stock prices and other financial variables.
• Engineering: Designing systems that can handle random variables and uncertainties.
• Computer Science: Developing algorithms that can handle random inputs and outputs.

Q: Can you provide more resources for learning about this topic?

A: Yes, there are many resources available for learning about this topic, including:

• Books: "Probability and Statistics: A Course for Graduate Students" by David R. Brillinger, "Probability Theory: The Logic of Science" by E. T. Jaynes, and "Random Processes: A First Course" by I. I. Gikhman and A. V. Skorokhod.
• Online Courses: "Probability and Statistics" on Coursera, "Probability Theory" on edX, and "Random Processes" on Udemy.
• Research Papers: Search for papers on arXiv, ResearchGate, and Academia.edu using keywords like "expected value", "highest variable so far", and "random variables".

Conclusion

In this article, we have answered some frequently asked questions about the expected value of the highest variable so far in a sequence of random variables. We hope that this article has provided valuable insights and information for readers who are interested in this topic.