Question

Let \(X_{1}, X_{2}, X_{3}\) be iid with common pdf \(f(x)=\exp (-x), 0

Short answer

The answers for the probabilities are (a) \(0.5\) and (b) \(1/6\).

Step-by-step solution

Analyze Part (a)

We are asked to find \(P\left(X_{1}<X_{2} \mid X_{1}<2 X_{2}\right)\). From the memoryless property of the exponential distribution, this is equivalent to \(P\left(X_{1}<X_{2}\right)\). Since \(X_1\) and \(X_2\) are iid, \(P\left(X_{1}<X_{2}\right) = P\left(X_{2}<X_{1}\right)\). Thus, it must be that the probability is \(0.5\).

Analyze Part (b)

Now, we want to find \(P\left(X_{1}<X_{2}<X_{3} \mid X_{3}<1\right)\). Since \(X_{1}\), \(X_{2}\), and \(X_{3}\) are independent, the probability of the three independent events can be obtained by integration over their joint pdf. The joint pdf \(f(x_1, x_2, x_3)\) of three exponential random variables is given by \[f(x_1, x_2, x_3) = f(x_1) f(x_2) f(x_3) = \exp(-x_1) \exp(-x_2) \exp(-x_3)\] if \(0 < x_1, x_2, x_3 < \infty\), and zero elsewhere.

Apply the Integral

The required probability then is \[P\left(X_{1}<X_{2}<X_{3} \mid X_{3}<1\right) = \int_{x_3=0}^{1} \int_{x_2=0}^{x_3} \int_{x_1=0}^{x_2} \exp(-x_1) \exp(-x_2) \exp(-x_3) dx_1 dx_2 dx_3\] After performing the integral calculations, the result is \(1/6\).