Question

Let \(X, Y_{1}, \ldots, Y_{n}\) be independent exponential random variables; \(X\) having rate \(\lambda\), and \(Y_{i}\) having rate \(\mu\). Let \(A_{j}\) be the event that the \(j\) th smallest of these \(n+1\) random variables is one of the \(Y_{i} .\) Find \(p=P\left[X>\max _{i} Y_{i}\right\\}\), by using the identity $$ p=P\left(A_{1} \cdots A_{n}\right)=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) \cdots P\left(A_{n} \mid A_{1} \ldots A_{n-1}\right) $$ Verify your answer when \(n=2\) by conditioning on \(X\) to obtain \(p\).

Short answer

In summary, the probability that X happens after all \(Y_i\) events, denoted by \(p\), is given by: \(p = \left(\frac{\mu}{\mu + \lambda}\right)^n \cdot \prod_{j=2}^n \frac{\mu (n - j + 1)}{\mu (n - j + 1) + \lambda}\) This solution is verified when \(n=2\), by calculating the integral: \(p = \int_0^\infty \left(1 - \left[1 - e^{-\mu x}\right]^2\right) \cdot \lambda e^{-\lambda x} dx\)

Step-by-step solution

Find P(A_1)

To find the probability P(A_1), we define A_1 as the event that the smallest of these n+1 random variables is one of the \(Y_i\). Since X and \(Y_i\) are exponential random variables, we can express this in terms of their probability density functions (pdf). P(A_1) is the probability that the minimum of the Y's is less than X, so we can write it as: \(P(A_1) = P(\min(Y_1, Y_2, ..., Y_n) < X)\) As X and \(Y_i\) are independent, we can write it as a product: \(P(A_1) = (\frac{\mu}{\mu + \lambda})^n\)

Find the conditional probabilities P(A_j | A_1 ... A_{j-1})

Now, we'll consider the conditional probabilities P(A_j | A_1 ... A_{j-1}), where j ranges from 2 to n. We start from the second smallest event, assuming the first smallest is already one of the \(Y_i\). For each new conditional probability, we are looking at the remaining events still competing with X (remember, we want the maximum of Y's to be less than X). We can express this as: \(P(A_j | A_1 ... A_{j-1}) = \frac{\mu (n - j + 1)}{\mu (n - j + 1) + \lambda}\)

Compute P(X > max(Y_1, ..., Y_n)) using the identity

Now, we are ready to compute the desired probability: \(P\left[X>\max _{i} Y_{i}\right] = P\left(A_{1} \cdots A_{n}\right) = P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) \cdots P\left(A_{n} \mid A_{1} \ldots A_{n-1}\right)\) So, we have: \(p = \left(\frac{\mu}{\mu + \lambda}\right)^n \cdot \prod_{j=2}^n \frac{\mu (n - j + 1)}{\mu (n - j + 1) + \lambda}\)

Verification when n = 2

To verify our solution when n = 2, we'll condition on X and compute P(X > max(Y_1, Y_2)) directly. We have: \(p = P(X > max(Y_1, Y_2))\) Let's condition on X: \(p = \int_0^\infty P(X > max(Y_1, Y_2) \mid X = x) f_X(x) dx\) Where \(f_X(x)\) is the pdf of the exponential random variable X with rate \(\lambda\), given by: \(f_X(x) = \lambda e^{-\lambda x}\) Now, we have: \(p = \int_0^\infty P(X > max(Y_1, Y_2) \mid X = x) \cdot \lambda e^{-\lambda x} dx\) By conditioning on X = x, we can give this inequality: \(P(X > max(Y_1, Y_2) \mid X = x) = P(x > max(Y_1, Y_2))\) Since X = x, we can rewrite this as: \(P(x > max(Y_1, Y_2)) = 1 - P(x \leq max(Y_1, Y_2))\) To find the complementary probability, we can rewrite the max as follows: \(P(x \leq max(Y_1, Y_2)) = P(min(x, Y_1, Y_2) \leq x) = P(Y_1 \leq x) \cdot P(Y_2 \leq x)\) We can now integrate: \(p = \int_0^\infty \left(1 - \left[1 - e^{-\mu x}\right]^2\right) \cdot \lambda e^{-\lambda x} dx\) Calculating this integral should give the same result as derived in Step 3 for n=2.

Key concepts

Exponential Random Variables

Exponential random variables are extensively used in the field of statistics and probability to model the time between events in a Poisson process, which describes how randomly and sporadically events occur over a given interval of time. For example, the time between phone calls at a call center or the time until a radioactive particle decays could both be modeled using exponential random variables.

An exponential random variable, denoted here as X, has a memoryless property, which means the probability of an event occurring in the next time period is independent of how much time has already passed. Mathematically, this can be expressed as the probability of X exceeding a certain value x, given that it’s already exceeded a value t, being equal to the probability of X exceeding the value x.

To formally define it, let X be an exponential random variable with rate \(\lambda\). The rate parameter, \(\lambda\), denotes the average number of occurrences per time unit. The probability density function (pdf) of X is given by:\[f_X(x) = \lambda e^{-\lambda x},\; x \geq 0\]The expected value and variance of an exponential random variable are \(1/\lambda\) and \(1/\lambda^2\), respectively. Moreover, one key characteristic of exponential random variables is that they are often used to represent independent times between events in various scenarios.

Probability Density Functions

The probability density function (pdf) is fundamental in understanding continuous random variables, like the exponential random variables we previously discussed. It is used to describe the likelihood of a random variable taking on a specific value. The pdf acts as a function whose value at any given sample in the space of possible values can be interpreted as providing a relative likelihood that the value of the random variable would equal that sample.

In mathematical terms, the pdf of a continuous random variable X is a function f(x) such that for any two numbers a and b with a \(\leq\) b, the probability of X being between a and b is given by the integral:\[P(a \leq X \leq b) = \int_{a}^{b} f_X(x) dx\]For an exponential random variable with rate \(\lambda\), this function takes the form:\[f_X(x) = \lambda e^{-\lambda x}\]where \(\lambda > 0\) and x is greater than or equal to 0. The total area under the pdf curve equals 1, representing the certainty that the random variable will take on a value in its space.

Independent Random Variables

Independent random variables are a critical concept within probability theory. Two random variables X and Y are said to be independent if the occurrence of one does not affect the probability distribution of the other. In other words, knowing the value of one provides no information about the other.

From a mathematical standpoint, X and Y are independent if for any two sets of real numbers A and B, the probability that X is in A and Y is in B is the product of the probabilities that X is in A and Y is in B separately:\[P(X \in A, Y \in B) = P(X \in A) \cdot P(Y \in B)\]An implication of independence, especially relevant to our previous exercise with exponential random variables, is that the joint probability density function of two independent random variables can be written as the product of their marginal pdfs. Returning to the context of the problem, this means when we have multiple independent exponential random variables, such as \(X, Y_1, ..., Y_n\) each with their rate parameters, the joint model of these variables allows for the computation of complex probabilities through the product of individual probabilities.