Problem 5 - Calculus for Biology and Medicine

Question

Let \(X\) be a continuous random variable with density function \(f(x)=\left\\{\begin{array}{cl}2 e^{-2 x} & \text { for } x>0 \\ 0 & \text { for } x \leq 0\end{array}\right.\) Find \(E(X)\) and \(\operatorname{var}(X)\)

Step-by-Step Solution

Answer

\(E(X) = 0.5\), \(\operatorname{var}(X) = 0.25\).

Step 1: Identify the probability density function (pdf)

The given probability density function for the random variable \(X\) is \(f(x)=2e^{-2x}\) for \(x > 0\) and \(0\) for \(x \leq 0\). This represents the density function of an exponential distribution with rate parameter \(\lambda = 2\).

Step 2: Calculate the expected value \(E(X)\)

For an exponential distribution with rate \(\lambda\), the expected value \(E(X)\) is given by \(\frac{1}{\lambda}\). Here, \(\lambda = 2\), so:\[E(X) = \frac{1}{2} = 0.5\]

Step 3: Calculate the variance \(\operatorname{var}(X)\)

The variance of an exponential distribution with rate \(\lambda\) is \(\frac{1}{\lambda^2}\). Therefore, given \(\lambda = 2\):\[\operatorname{var}(X) = \frac{1}{2^2} = \frac{1}{4}\]

Step 4: Provide final results for \(E(X)\) and \(\operatorname{var}(X)\)

The expected value of \(X\) is \(E(X) = 0.5\) and the variance of \(X\) is \(\operatorname{var}(X) = \frac{1}{4}\).

Key Concepts

Expected Value of an Exponential Distribution

The expected value, often denoted as \(E(X)\), represents the average or mean value of a random variable when repeated trials are conducted. In the specific context of exponential distributions, the expected value is a straightforward calculation. The exponential distribution is defined by a single rate parameter, \(\lambda\). This parameter influences the expected time between events in a Poisson process, which the exponential distribution models.

For an exponential distribution with rate parameter \(\lambda\), the formula for computing the expected value is:

• \(E(X) = \frac{1}{\lambda}\)

In the given problem, the rate parameter \(\lambda\) is equal to 2, hence the expected value can be computed as \(\frac{1}{2} = 0.5\). This means, on average, the event occurs every 0.5 units of time.

The beautiful simplicity of this formula makes exponential distributions particularly appealing for modeling certain real-world processes. It's a key metric in determining how long, on average, you should expect until an event occurs.

Variance of an Exponential Distribution

Variance measures the spread of a set of values in a random variable. It indicates how much the values deviate from the expected value or mean on average. For exponential distributions, the variance can also be calculated using the rate parameter \(\lambda\), and is derived from the underlying properties of the distribution.

To find the variance for an exponential distribution, you can use the formula:

• \(\text{var}(X) = \frac{1}{\lambda^2}\)

In this case, since \(\lambda = 2\), the variance is \(\frac{1}{2^2} = \frac{1}{4}\). This result suggests that while the average waiting time between events is 0.5 units, there is some variability in the actual times which can be succinctly characterized by this variance value.

The variance helps us understand the predictability of the wait time between certain events. In contexts like waiting for service or between phone calls in a call center, understanding variability is crucial for effective resource management.

Probability Density Function of an Exponential Distribution

A probability density function (PDF) describes the likelihood of a random variable to take on a particular value. In the case of continuous random variables like those following an exponential distribution, the PDF provides this likelihood for each possible outcome.

For an exponential distribution with rate parameter \(\lambda\), the PDF is given by the equation:

• \(f(x) = \lambda e^{-\lambda x}\) for \(x > 0\)
• \(f(x) = 0\) for \(x \leq 0\)

This function indicates that the probability rapidly decreases as \(x\) increases, which aligns with the idea that shorter intervals between events are more probable than longer ones.

In our problem, \(\lambda = 2\), so the PDF becomes \(f(x) = 2e^{-2x}\) for \(x > 0\). This specific form highlights the stochastic nature of processes modeled by the exponential distribution, such as time between arrivals in queueing models. The exponential distribution's simplicity in expressing event time probabilities makes it a foundational tool in reliability engineering and risk management.