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 and \(\operatorname{var}(X)\) = 0.25.

Step 1: Understanding the Problem

We need to find the expected value, \(E(X)\), and the variance, \(\operatorname{var}(X)\), of a continuous random variable \(X\) with a given probability density function (pdf). This pdf is \(f(x) = 2e^{-2x}\) for \(x > 0\) and 0 otherwise, which suggests \(X\) follows an exponential distribution with rate \(\lambda = 2\).

Step 2: Determine Expected Value (E(X))

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

Step 3: Determine Variance (Var(X))

The variance of an exponential distribution with rate \(\lambda\) is \(\operatorname{var}(X) = \frac{1}{\lambda^2}\). Given \(\lambda = 2\), we calculate:\[ \operatorname{var}(X) = \frac{1}{2^2} = \frac{1}{4} = 0.25 \]

Step 4: Final Step: Summarize the Results

Now that we have computed both, the expected value \(E(X)\) of the distribution is 0.5 and the variance \(\operatorname{var}(X)\) is 0.25.

Key Concepts

Expected Value

The concept of expected value is fundamental in understanding random variables. Think of expected value as the average outcome you'd anticipate if you were to repeat an experiment an infinite number of times.
In the context of a continuous random variable following an exponential distribution, the expected value helps describe the central tendency of the distribution.
For an exponential distribution with rate parameter \( \lambda \), the expected value \( E(X) \) is calculated using the formula:

• \( E(X) = \frac{1}{\lambda} \)

In our example, with \( \lambda = 2 \), this leads to:

• \( E(X) = \frac{1}{2} = 0.5 \)

This means if we observe the process described by this exponential random variable repeatedly, we expect the average value to be about 0.5.

Variance

Variance provides a measure of how spread out the values of a random variable are from the expected value. In simpler terms, it gives us an idea of the variability or uncertainty of the outcomes we might expect.
For a continuous random variable like one with an exponential distribution, variance quantifies the extent to which we might expect values to deviate from the mean.
When dealing with an exponential distribution, the variance is calculated using:

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

Here, with a rate \( \lambda = 2 \), the variance works out to be:

• \( \operatorname{var}(X) = \frac{1}{2^2} = \frac{1}{4} = 0.25 \)

This tells us that, relative to the mean, values can be expected to vary with a degree of uncertainty of about 0.25, reflecting the extent of the spread.

Continuous Random Variable

A continuous random variable is one which can take on an infinite number of possible values. Unlike a discrete random variable, which is only defined at specific points, continuous random variables are defined over an interval.
This means for any given range of values, the probability of the variable falling within that range is determined by integrating its probability density function (pdf).
For instance, in our exercise the random variable \( X \) has a pdf given by:

• \( f(x)=2 e^{-2 x} \) for \( x>0 \)
• \( f(x)=0 \) for \( x \leq 0 \)

Such a pdf indicates that \( X \) follows an exponential distribution. This distribution is optimal for modeling time until an event occurs, such as failure times or waiting periods. Therefore, the exponential distribution is a prime example of a continuous random variable.