Question

Let \(h(x)=P\left(\sum_{i=1}^{T} X_{i}>x\right\\}\) where \(X_{1}, X_{2}, \ldots\) are independent random variables having distribution function \(F_{e}\) and \(T\) is independent of the \(X_{i}\) and has probability mass function \(P[T=n\\}=\rho^{n}(1-\rho), n \geqslant 0 .\) Show that \(h(x)\) satisfies Equation (7.53). Hint: Start by conditioning on whether \(T=0\) or \(T>0\).

Short answer

In summary, we have shown that the given function \(h(x)\) can be expressed as: \[h(x) = \left(P(0>x)\cdot(1-\rho)\right) + \left(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\cdot1\right).\] This is done by conditioning on \(T=0\) or \(T>0\) and using the probability mass function and distribution function provided. Although we couldn't prove the relation to the unspecified Equation (7.53), we demonstrated a step-by-step approach to breaking down the exercise based on the given hint.

Step-by-step solution

Calculate the Probabilities for \(T=0\) and \(T>0\)

To calculate \(P(T=0)\) and \(P(T>0)\), use the given probability mass function \(P[T=n]=\rho^n(1-\rho), n \geqslant 0\). We have: 1. For \(n=0\): \[P(T=0) = \rho^0(1-\rho) = 1-\rho.\] 2. For \(n>0\): \[P(T>0) = \sum_{n=1}^{\infty} \rho^n(1-\rho) = 1-\rho + \rho^2(1-\rho) + \rho^3(1-\rho) + \cdots .\] This is an infinite geometric series with the first term \(a = (1-\rho)\) and common ratio \(r=\rho\). The sum formula for an infinite geometric series is: \[\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}.\] So with this: \[P(T>0) = \frac{1-\rho}{1-\rho} =1.\]

Calculate Independent Probabilities

Now, let's calculate \(P\left(\sum_{i=1}^{T} X_{i} > x | T = 0\right)\) and \(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\): 1. If \(T=0\), there is no random variable \(X_i\), so the sum is equal to 0: \[P\left(\sum_{i=1}^{T} X_{i} > x | T = 0\right) = P(0>x).\] 2. If \(T>0\), the result can be obtained via the given distribution function \(F_{e}\): \[P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right).\]

Calculate the Function \(h(x)\)

Finally, substitute the calculated probabilities in the formula of \(h(x)\): \[h(x) = \left(P(0>x)\cdot(1-\rho)\right) + \left(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\cdot1\right).\] Since we don't have the exact equation (7.53), we could not prove the direct relation. However, we have demonstrated how to break down the exercise by following the given hint and considering the conditions on \(T=0\) and \(T>0\).

Key concepts

Infinite Geometric Series

Understanding an infinite geometric series is key to solving many problems in probability and other areas of mathematics. An infinite geometric series is a sum of the form \(\sum_{n=1}^{\infty} ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio. For the series to converge, the common ratio \(r\) must satisfy \(|r| < 1\). The sum of an infinite geometric series can be found using the formula \(S = \frac{a}{1-r}\) if \(r\) is strictly between -1 and 1.

In our exercise, we have a probability mass function that references such a series. When calculating the probability that \(T > 0\), we've identified an infinite geometric series with the first term \(a = (1-\rho)\) and common ratio \(r = \rho\). Using the sum formula for an infinite geometric series, we can easily find that \(P(T>0)\) equals 1. This is a critical step in probability calculations, especially when dealing with random processes over an indefinite number of trials or time periods.

Random Variables

A random variable is a valuable concept in statistics and probability theory. It's essentially a variable whose value is subject to variations due to chance. Random variables can be classified into two categories: discrete and continuous. Discrete random variables take on a countable number of distinct values. Continuous random variables, on the other hand, can take on any value within a given range.

In the context of our exercise, \(X_1, X_2, ..., X_T\) are independent random variables with distribution function \(F_e\), and these variables represent distinct outcomes of a probabilistic event. Independence means that the result of one variable does not influence the others. The sum \(\sum_{i=1}^{T} X_{i}\) is then a new random variable, and its behavior is essential when calculating the function \(h(x)\), which represents a specific probability regarding the sum of these variables exceeding a certain value \(x\).

Distribution Function

The distribution function, also known as the cumulative distribution function (CDF), is a fundamental concept in probability. It describes the probability that a real-valued random variable \(X\) will have a value less than or equal to \(x\). Mathematically, for a given value of \(x\), the distribution function \(F(x)\) is calculated as \(F(x) = P(X \leq x)\).

In our exercise, \(F_e\) is the distribution function of the random variables \(X_i\). This function helps to find probabilities related to individual random variables out of a sequence of independent random variables. By using this distribution function, we can compute probabilities like \(P\left(\sum_{i=1}^{T} X_{i} > x | T > 0\right)\), which is a necessary element in deriving our desired function \(h(x)\). The understanding of CDF is crucial in determining probabilities associated with sums of random variables, as it allows for assessing the accumulated likelihoods up to a certain point.