Question

Let \(X\) and \(Y\) be independent exponential random variables with respective rates \(\lambda\) and \(\mu\). Let \(M=\min (X, Y)\). Find (a) \(E[M X \mid M=X]\) (b) \(E[M X \mid M=Y]\) (c) \(\operatorname{Cov}(X, M)\)

Short answer

(a) \(E[M X \mid M=X] = \frac{2}{\lambda^2}\) (b) \(E[M X \mid M=Y] = \frac{1}{\lambda + \mu}\) (c) \(\operatorname{Cov}(X, M) = \frac{-\mu}{\lambda^2(\lambda + \mu)}\)

Step-by-step solution

Define the PDFs of X and Y

Since X and Y are exponential random variables with rates \(\lambda\) and \(\mu\) respectively, their PDFs are given by: $$f_X(x) = \lambda e^{-\lambda x}, x > 0$$ $$f_Y(y) = \mu e^{-\mu y}, y > 0$$ Since X and Y are independent, the joint PDF is given by the product of their individual PDFs: $$f_{XY}(x,y) = f_X(x) f_Y(y) = \lambda e^{-\lambda x} \mu e^{-\mu y}$$

Compute \(E[X^2 \mid M=X]\)

To compute the conditional expectation, we first need to find the conditional PDF \(f_{X^2 | M=X}\). Since \(M=X\) when \(X<Y\), we only need to consider the case when \(x<y\). Hence, the conditional PDF is given by: $$f_{X^2 | M=X}(x^2) = \int_{x}^{\infty} f_{XY}(x,y) \, dy = \int_{x}^{\infty} \lambda e^{-\lambda x} \mu e^{-\mu y} \, dy$$ Now we compute the conditional expectation: $$E[X^2 \mid M=X] = \int_{0}^{\infty} x^2 \, f_{X^2 | M=X}(x^2) \, dx$$ After evaluating the integral, we get: $$E[X^2 \mid M=X] = \frac{2}{\lambda^2}$$

Compute \(E[XY \mid M=Y]\)

Similarly, to compute this conditional expectation, we first need to find the conditional PDF \(f_{XY | M=Y}\). Since \(M=Y\) when \(Y<X\), we only need to consider the case when \(y<x\). Hence, the conditional PDF is given by: $$f_{XY | M=Y}(xy) = \int_{y}^{\infty} f_{XY}(x,y) \, dx = \int_{y}^{\infty} \lambda e^{-\lambda x} \mu e^{-\mu y} \, dx$$ Now we compute the conditional expectation: $$E[XY \mid M=Y] = \int_{0}^{\infty} \int_{y}^{\infty} xy \, f_{XY | M=Y}(xy) \, dx \, dy$$ After evaluating the integral, we get: $$E[XY \mid M=Y] = \frac{1}{\lambda + \mu}$$

Compute \(\operatorname{Cov}(X, M)\)

To compute the covariance, we need to find \(E[XM]\), \(E[X]\), and \(E[M]\). First, note that since \(M = \min(X,Y)\), we have the PDF of \(M\) as: $$f_M(m) = \lambda e^{-\lambda m} + \mu e^{-\mu m}$$ Now compute the expectations: $$E[X] = \int_{0}^{\infty} x \, f_X(x) \, dx = \frac{1}{\lambda}$$ $$E[M] = \int_{0}^{\infty} m \, f_M(m) \, dm = \frac{1}{\lambda + \mu}$$ $$E[XM] = \int_{0}^{\infty} \int_{0}^{\infty} xm \, f_{XY}(x,m) \, dx \, dm = \frac{\lambda + 2 \mu}{\lambda^2 (\lambda + \mu)^2}$$ Finally, compute the covariance: $$\operatorname{Cov}(X, M) = E[XM] - E[X]E[M] = \frac{-\mu}{\lambda^2(\lambda + \mu)}$$ To summarize the results, we have: (a) \(E[M X \mid M=X] = \frac{2}{\lambda^2}\) (b) \(E[M X \mid M=Y] = \frac{1}{\lambda + \mu}\) (c) \(\operatorname{Cov}(X, M) = \frac{-\mu}{\lambda^2(\lambda + \mu)}\)