91Ó°ÊÓ

The PDF of an exponential random variable \(X_i\) with parameter \(\lambda_i\) is given by: \[f_{X_i}(x) = \lambda_i e^{-\lambda_i x}, \hspace{5mm} x > 0, \hspace{2mm} i \in \{1, 2\}\]

We have the transformation: \(Z = \frac{X_1}{X_2}\). Let's find the inverse transformations: \[X_1 = X_2 Z\] \[X_2 = X_1 / Z\] Now, let's find the Jacobian of the transformation: \[\frac{\partial(x_1, x_2)}{\partial(z, x_2)} = \begin{vmatrix} x_2 & -\frac{x_1}{z^2} \\ z & 0 \\ \end{vmatrix} = -x_1\] We know that the joint PDF of two independent random variables is the product of their individual PDFs. Therefore, the joint PDF \(f_{X_1, X_2}(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2) = \lambda_1 e^{-\lambda_1 x_1} \lambda_2 e^{-\lambda_2 x_2}\). Thus, we can find the PDF of \(Z\) using the transformation technique: \[f_Z(z) = \int_{0}^{\infty} |J_\text{transform}| \cdot f_{X_1, X_2}(X_1(z, x_2), X_2(z, x_2)) dx_2\]

Now, we substitute the values of the Jacobian and the joint PDF into the equation above and compute the integral: \[f_Z(z) = \int_{0}^{\infty} |(-x_1)| \cdot (\lambda_1 e^{-\lambda_1 (x_2z)} \lambda_2 e^{-\lambda_2 x_2}) dx_2\] \[f_Z(z) = \lambda_1 \lambda_2 \int_{0}^{\infty} x_1 e^{-\lambda_1 (x_2z)} e^{-\lambda_2 x_2} dx_2\] We will substitute \(x_1 = x_2/z\), and thus, \[f_Z(z) = \frac{\lambda_1 \lambda_2}{z} \int_{0}^{\infty} x_2 e^{-\lambda_1 (x_2z)} e^{-\lambda_2 x_2} dx_2\] Now, integrate by parts, with \(u = x_2\) and \(dv = e^{-\lambda_1 (x_2z) -\lambda_2 x_2} dx_2\): \[f_Z(z) = \frac{\lambda_1 \lambda_2}{\lambda_1 z + \lambda_2}\, \frac{1}{z}\]

To find the probability that \(X_1\) is less than \(X_2\), we will find the CDF of \(Z\): \[P(Z<1) = F_Z(1) = \int_{0}^1 f_Z(z) dz\] \[F_Z(1) = \int_{0}^1 \frac{\lambda_1 \lambda_2}{\lambda_1 z + \lambda_2}\, \frac{1}{z} dz\] Now, we will perform the substitution: \(u = \lambda_1 z + \lambda_2\), which implies \(du = \lambda_1 dz\). Then, the limits of integration will change as follows: When \(z = 0\), \(u = \lambda_2\) When \(z = 1\), \(u = \lambda_1 + \lambda_2\) \[F_Z(1) = \int_{\lambda_2}^{\lambda_1 + \lambda_2} \frac{1}{u} du\] \[F_Z(1) = \ln\Big(\frac{\lambda_1+\lambda_2}{\lambda_2}\Big)\] Hence, the required probability is: \[P\{X_1 < X_2\} = P(Z<1) = F_Z(1) = \ln\Big(\frac{\lambda_1+\lambda_2}{\lambda_2}\Big)\]