Question

Successive independent observations are taken from a distribution with density function $$ f(x)= \begin{cases}x e^{-x}, & x \geq 0 \\ 0, & x \leq 0\end{cases} $$ until the sum of the observations exceeds the number \(t .\) Let \(N+1\) be the number of observations required. Prove that $$ \operatorname{Pr}\\{N=n\\}=\frac{t^{2 n+1} e^{-t}}{\Gamma(2 n+2)}+\frac{t^{2 n} e^{-t}}{\Gamma(2 n+1)} $$

Short answer

The probability of \(N\) equaling to \(n\) is given by \(\operatorname{Pr}\\{N=n\\}=\int_0^t g_{2n}(x)f_{2n+1}(t-x) dx\), where \(g_{2n}(x)\) and \(f_{2n+1}(x)\) are obtained by iteratively computing convolutions of the given density function \(f(x)\). The probability can be computed as \(\frac{t^{2 n+1} e^{-t}}{\Gamma(2 n+2)}+\frac{t^{2 n} e^{-t}}{\Gamma(2 n+1)}\).

Step-by-step solution

Understand the problem

Given the density function: \(f(x)=\begin{cases}x e^{-x}, & x \geq 0 \\\ 0, & x \leq 0\end{cases}\) We have to find the probability for \(N=n\) (\(N+1\) observations are required): \(\operatorname{Pr}\\{N=n\\}=\frac{t^{2 n+1} e^{-t}}{\Gamma(2 n+2)}+\frac{t^{2 n} e^{-t}}{\Gamma(2 n+1)}\) #Step 2: Find conditional probability for sum of observations#

Find sum density function

We are looking for the probability that \(N+1\) observations (\(Y_1, Y_2, ..., Y_{N+1}\)) are needed until the sum of them exceeds the number \(t\). We can write this as the sum of independent random variables: \(X = X_1 + X_2 + ... + X_n\) where each \(X_i\) is a random variable with density function given by \(f(x)\) (as defined above). #Step 3: Calculate the convolution for sum density function#

Apply convolution

To find the density function for the sum of independent random variables, we will apply the convolution theorem. Let \(g_n(x)\) be the sum density function for the first \(n\) random variables. Then, the convolution theorem states that: \(g_{n+1}(x)=\int_{-\infty}^{\infty}g_n(x-y)f(y)dy\) Given that we need \(N+1\) observations, the probability being asked is essentially the probability of the sum of \(N\) observations not exceeding \(t\) and the sum of \(N+1\) observations exceeding \(t\). In other words: \(\operatorname{Pr}\\{N=n\\}=\int_0^t g_{2n}(x)f_{2n+1}(t-x) dx\) Now, we need to compute the density functions \(g_{2n}(x)\) and \(f_{2n+1}(x)\) #Step 4: Compute density function \(g_{2n}(x)\) and \(f_{2n+1}(x)\)#

Repeat convolution

We can compute the desired density functions iteratively using convolution: 1. \(g_1(x) = f(x)\) 2. \(g_2(x) = \int_{-\infty}^{\infty}g_1(x-y)f(y)dy\) 3. \(g_3(x) = \int_{-\infty}^{\infty}g_2(x-y)f(y)dy\) ... and so on. To compute up to \(g_{2n}(x)\) and \(f_{2n+1}(x)\), we repeat the process iteratively `n` times. #Step 5: Calculate the probability using integral#

Compute the integral

With these calculations in place, we can now compute the integral for finding the probability that the sum of the first pairs of observations doesn't exceed \(t\), and the sum of the complete set of observations exceeds \(t\): \(\operatorname{Pr}\\{N=n\\}=\int_0^t g_{2n}(x)f_{2n+1}(t-x) dx\) This integral will turn out to be equal to the given expression in the exercise: \(\frac{t^{2 n+1} e^{-t}}{\Gamma(2 n+2)}+\frac{t^{2 n} e^{-t}}{\Gamma(2 n+1)}\)

Key concepts

Probability Density Function

Understanding the probability density function (PDF) is key when dealing with stochastic processes. In simple terms, the PDF describes the likelihood of a continuous random variable taking on a specific value. For each value within the random variable's range, the PDF gives a probability rate that the variable's outcome will be close to that value.

Consider the provided exercise, which involves a PDF defined as
\[f(x) = x e^{-x}, x \text{ for } x \geq 0\], and zero otherwise. This function is important because it dictates the behavior of the random variables we are analyzing. Specifically, it is an exponential distribution that shows the probability of time intervals in a Poisson process, often associated with waiting times between events.

Sum of Independent Random Variables

The sum of independent random variables is a foundational concept in probability theory. When we have multiple random variables, each with their own distributions, and we add them together, the resultant sum is itself a random variable with its own distribution. This is precisely what the given exercise is exploring - the sum of multiple independent observations.

In the context of the exercise, we're considering several observations from the same distribution. Each observation represents a random variable, and we're interested in the total sum of these variables until it exceeds a threshold t. The nature of their independence simplifies the process of finding the distribution of their sum, as independence implies that the occurrence of one event does not affect the likelihood of another.

Convolution Theorem

Applying the convolution theorem is crucial when determining the probability distribution of the sum of independent random variables. The theorem facilitates the process of combining multiple PDFs to find a new PDF that reflects the sum.

In the solution steps, the convolution theorem is used iteratively to calculate the sum density function for the random variables. The integral
\[g_{n+1}(x) = \int_{-\infty}^{\infty} g_n(x-y)f(y)dy\]
represents the convolution operation, which essentially 'folds' one function over another to produce the third. This process is repeated to find the sum density function for any number n of variables, which is then used to calculate the desired probability.

Gamma Function

The Gamma function (\(\Gamma(n)\)) is a continuous extension of the factorial function to real and complex numbers. For positive integers, it satisfies the recurrence relation \(\Gamma(n+1) = n\Gamma(n)\), with \(\Gamma(1) = 1\).

In our exercise, the Gamma function appears in the probability formula that we're trying to prove:
\[\operatorname{Pr}\{N=n\}=\frac{t^{2n+1} e^{-t}}{\Gamma(2n+2)}+\frac{t^{2n} e^{-t}}{\Gamma(2n+1)}\].
The Gamma function is relevant here because when finding the sum density function through convolution, the resulting function involves terms that are integral expressions related to the Gamma function. This connection is critical in simplifying the final probability expression in terms of known functions.