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Chapter 2: Problem 6

Let \(f(x, y)=e^{-x-y}, 0

Short Answer

Expert verified
The probabilities are \(P(Z \leq 0) = 0\) and \(P(Z \leq 6) = 1 - e^{-6}\). The pdf of \(Z\) is a gamma distribution with shape parameter \(2\) and scale parameter \(1\).

Step by step solution

01

Joint PDF Calculation

The first step is to interpret the information given in the problem. \(f(x,y)\) is a joint probability density function (pdf) with \(x > 0\) and \(y > 0\). In this case, the function is \(e^{-x-y}\), which is a product of \(e^{-x}\) and \(e^{-y}\) suggesting that \(X\) and \(Y\) are independent.
02

Integrate over PDF

Next, in order to compute the probabilities, we need to integrate over the joint pdf. Because \(X\) and \(Y\) are independent, this is essentially an integral of the joint pdf over the region of integration. So for \(P(Z \leq 0)\), \(Z = X + Y\), and since neither \(X\) or \(Y\) can be negative, we can say directly that \(P(Z \leq 0) = 0\). For \(P(Z \leq 6)\), we would integrate from \(0\) to \(6\) using the joint pdf.
03

Compute the PDF of Z

Finally, to find the probability density function of \(Z\), one must note that when two independent variables are added together, their PDFs convolve. This gives \(h(z) = \int_{-\infty}^{+\infty} f(x) f(z - x) dx\). Since \(X\) and \(Y\) have exponential distributions, their sum \(Z\) will have a gamma distribution. The resulting pdf corresponds to a gamma distribution with shape parameter \(2\) and scale parameter \(1\).

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