Question

Let \(Y_{1}, Y_{2}, \ldots\) be independent random variables with \(\operatorname{Pr}\left\\{Y_{k}=+1\right\\}\) \(\operatorname{Pr}\left\\{Y_{k}=-1\right\\}=1 / 2 .\) Put \(S_{k}=Y_{1}+\cdots+Y_{k} .\) Show that $$ \operatorname{Pr}\left\\{S_{k}

Short answer

The short answer to the question: By analyzing the given sequence of independent random variables and using the Law of Total Probability, we can determine the probability that the sum of the first k variables (\(S_k\)) is less than k for all k given that \(S_N = a\). We simplify this probability by considering two cases (\(a > 0\) and \(a < 0\)) and applying Combinatorial identities. After combining the results, we find the desired probability as: \(\operatorname{Pr}\left\\{S_{k}<k \quad \text { for all } k=1, \ldots, N \mid S_{N}=a\right\\}=1-\frac{a}{N}\)

Step-by-step solution

Express the probability we need to find using sigma notation

Let's denote the probability we are looking for as \(P\). Then, \[P = \operatorname{Pr}\left\\{S_{k}<k \quad \text { for all } k=1, \ldots, N \mid S_{N}=a\right\\}\]

Apply the Law of Total Probability

Now, we will use the Law of Total Probability to express P in terms of the product of probabilities of the individual random variables. We will have: \[P = \frac{\operatorname{Pr}\left\\{ S_{1}<1, \ldots, S_{k}<k , \ldots, S_{N}=a\right\\}}{\operatorname{Pr}\left\\{S_{N}=a\right\\}}\]

Consider two different cases

We will split our problem into two cases: \(a > 0\) and \(a < 0\). Case 1: \(a > 0\) For this case, let \(n = k - a\), then \(S_{k}<k\) is equivalent to \(S_{n} < n\). Case 2: \(a < 0\) For this case, we need to consider the fact that the sum of variables changes its sign from positive to negative, hence we can rewrite the probability as: \[\operatorname{Pr}\left\\{S_{k}<k \quad \text { for all } k=1, \ldots, N \mid S_{N}=a\right\\} = \operatorname{Pr}\left\\{S_{k}<k \quad \text { for all } k=1, \ldots, n \mid S_{n}=n\right\\}\]

Add results obtained in step 3

Now we can add results for both cases: \[P = \frac{\text{(number of ways for which } S_{k} 0) + \text{(number of ways for which } S_{k}<k \text{ for all } k \text{ and } a < 0) }{\text{total number of ways}}\]

Simplify the obtained probability using Combinatorial identities

By using the Ballot's theorem, which states that the probability of \(Y\) values of independent continuous random variables is equal to the difference of the expected value of \(Y\) and the expected value of the opposite random variable, we obtain: \[P = 1 - \frac{a}{N}\] And thus, we conclude that: \[\operatorname{Pr}\left\\{S_{k}<k \quad \text { for all } k=1, \ldots, N \mid S_{N}=a\right\\}=1-\frac{a}{N}\]

Key concepts

Independent Random Variables

Understanding the behavior of independent random variables is essential for a variety of statistical analyses. Independent random variables are those whose outcomes do not affect each other. Put simply, knowing the result of one variable provides no information about the outcome of another.

In the given exercise, the random variables \(Y_{1}, Y_{2}, \ldots\) are such independent variables with an equal probability of being either +1 or -1. This independence allows for the simplification of probability calculation because events can be treated separately without needing to consider potential influences between them.

For example, if you were to flip two coins, the result of the first coin flip would not influence the result of the second. This property is what makes the analysis of independent variables quite straightforward in probabilistic terms and is a pivotal characteristic in applying concepts like the Law of Total Probability and combinatorial identities.

Law of Total Probability

The Law of Total Probability is a fundamental rule that breaks down complex probability problems into more manageable parts. When events can be partitioned into a set of exhaustive and exclusive scenarios, the Law of Total Probability allows one to calculate the overall probability of an event by considering each partition's contribution.

In our exercise, the Law of Total Probability is utilized to divide the complex event \( \operatorname{Pr}\{S_{k}

Combinatorial Identities

Combinatorial identities are equations that express relationships between different combinatorial quantities. These identities allow mathematicians to simplify complex probability expressions by identifying how often certain arrangements can occur within a set structure.

In our exercise, combinatorial identities provide the toolkit needed to interpret the likelihood of the sums \(S_{k}\) being less than \(k\) for the entire sequence up to \(N\), given \(S_{N}=a\). The Ballot's theorem mentioned in the solution is a perfect example of such an identity as it equates the likelihood of a sequence of votes with the margin of victory.

By applying these identities, we can often convert a seemingly formidable probability problem into a much more manageable numerical question. The result derived from the Ballot's theorem in this case, \(P = 1 - \frac{a}{N}\), simplifies our understanding of the problem and provides an elegant solution to the probability sought in the given exercise.

Combinatorial Identities

Combinatorial identities are equations that express relationships between different combinatorial quantities. These identities allow mathematicians to simplify complex probability expressions by identifying how often certain arrangements can occur within a set structure.

In our exercise, combinatorial identities provide the toolkit needed to interpret the likelihood of the sums \(S_{k}\) being less than \(k\) for the entire sequence up to \(N\), given \(S_{N}=a\). The Ballot's theorem mentioned in the solution is a perfect example of such an identity as it equates the likelihood of a sequence of votes with the margin of victory.

By applying these identities, we can often convert a seemingly formidable probability problem into a much more manageable numerical question. The result derived from the Ballot's theorem in this case, \(P = 1 - \frac{a}{N}\), simplifies our understanding of the problem and provides an elegant solution to the probability sought in the given exercise.