Problem 8 - A First Course in Probability

Question

The lifetime in hours of an electronic tube is a random variable having a probability density function given by $$ f(x)=x e^{-x} \quad x \geq 0 $$ Compute the expected lifetime of such a tube.

Step-by-Step Solution

Answer

The expected lifetime of the electronic tube, given the probability density function \(f(x) = xe^{-x}\) for \(x \geq 0\), is calculated as \(E(X) = 1\). Therefore, the expected lifetime of such a tube is 1 hour.

Step 1: Define the Expected Value Formula

The expected value (or mean) of a continuous random variable is given by the following formula: \[ E(X) = \int_{-\infty}^{\infty} x f(x) dx \] Here, \(X\) is the random variable representing the lifetime of the electronic tube, and \(f(x)\) is its PDF.

Step 2: Apply the Given PDF

We are given the PDF for the lifetime of the tube as \(f(x) = xe^{-x}\) for \(x \geq 0\). Since the lifetime \(x\) cannot be negative, we will integrate this function over the range \([0, \infty)\). The expected value formula becomes: \[ E(X) = \int_{0}^{\infty} x(xe^{-x}) dx \]

Step 3: Integrate the Function

To integrate the function, we will use integration by parts. Recall that integration by parts is given by: \[ \int u dv = uv - \int v du \] We will set \(u = x\) and \(dv = xe^{-x} dx\). Now we must find \(du\) and \(v\): - Differentiate \(u\) with respect to \(x\): \(du = dx\) - Integrate \(dv\) with respect to \(x\): \(v = -e^{-x}\) Applying integration by parts, we have: \[ E(X) = \int_{0}^{\infty} x(xe^{-x}) dx = -x e^{-x} \Big|_0^{\infty} - \int_{0}^{\infty} -e^{-x} dx \]

Step 4: Evaluate the First Integral

The first integral is a definite integral, so we will evaluate it at the bounds: \[ -x e^{-x} \Big|_0^{\infty} = \lim_{x\to\infty} (-x e^{-x}) - (-0\cdot e^0) = 0 \] We are left with: \[ E(X) = \int_{0}^{\infty} e^{-x} dx \]

Step 5: Evaluate the Remaining Integral

Now we need to evaluate the remaining integral: \[ E(X) = \int_{0}^{\infty} e^{-x} dx = -e^{-x} \Big|_0^{\infty} \] Evaluating the integral at the bounds, we have: \[ -e^{-x} \Big|_0^{\infty} = \lim_{x\to\infty} (- e^{-x}) - (-e^0) = 0 + 1 = 1 \]

Step 6: State the Expected Lifetime

The expected value (lifetime) of the electronic tube, as calculated from the PDF, is: \[ E(X) = 1 \] Thus, the expected lifetime of such an electronic tube is 1 hour.

Key Concepts

Expected Value

The expected value is a critical concept in probability and statistics. It represents the average or mean outcome you can expect when an experiment is repeated many times. For continuous random variables like the lifetime of an electronic tube, the expected value is calculated by integrating the product of the random variable and its probability density function (PDF) across all possible values.

The formula for expected value, denoted as \(E(X)\), in the continuous case is:

• \(E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \)

Here, \(f(x)\) is the PDF of the random variable \(X\). This integral essentially weights each outcome by its probability, summing them to find the expected or average value over the entire range.

Probability Density Function

A probability density function (PDF) is a mathematical function that describes the likelihood of a continuous random variable taking on a particular value. It is a cornerstone of continuous probability distributions, like those expressing the time until an event occurs.

The shape and behavior of the PDF determine the probability of different outcomes within defined intervals.

• In our example with the lifetimes of electronic tubes, the PDF is \( f(x) = x e^{-x} \) for \( x \geq 0 \).
• This specific function tells us how probable it is for a tube to have a particular lifetime, and longer lifetimes see an exponentially decreasing probability.

The area under the curve of the PDF between two points gives the probability that the random variable falls within that interval.

Integration by Parts

Integration by parts is a technique used to solve integrals of products of functions. It is particularly helpful when faced with integrals that cannot be easily evaluated using standard methods. This technique is a reflection of the product rule for differentiation and is essential in solving expected value problems with complex PDFs.

The integration by parts formula is:

• \(\int u \, dv = uv - \int v \, du\)

When applying this method, you identify parts of your function to treat as \(u\) and \(dv\), then differentiate \(u\) to get \(du\), and integrate \(dv\) to get \(v\). This was precisely how the expected value calculation for the electronic tube was approached, breaking down the complex integral into manageable parts.

Continuous Random Variable

A continuous random variable is one that can take an infinite number of possible values. These variables typically represent measurements such as time, length, and temperature, which are not countable in discrete units. The key aspect of continuous random variables is that their set of possible outcomes forms a continuum, often constrained within some bounds.

Examples include the amount of time until a light bulb burns out or the height of individuals in a population.

• The lifetime of an electronic tube is one such variable, measurable in hours, minutes, and even seconds, leading to its representation by a continuous PDF.
• For continuous random variables, the probability of it assuming a specific single value is practically zero; instead, we calculate probabilities over intervals.

These variables are often described by probability density functions which provide insight into their behavior across different ranges.