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Chapter 5: Problem 8

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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: \[ E(X) = \int_{0}^{\infty} x(xe^{-x}) dx \]
03

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: \[ E(X) = \int_{0}^{\infty} x(xe^{-x}) dx = -x e^{-x} \Big|_0^{\infty} - \int_{0}^{\infty} -e^{-x} dx \]
04

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: \[ E(X) = \int_{0}^{\infty} e^{-x} dx \]
05

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: \[ -e^{-x} \Big|_0^{\infty} = \lim_{x\to\infty} (- e^{-x}) - (-e^0) = 0 + 1 = 1 \]
06

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.

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