Exercise 10 p303

Let X1, X2, . . . , Xn be n independent random variables each of which has an exponential density with mean μ. Let M be the minimum value of the Xj . Show that the density for M is exponential with mean μ/n. Hint: Use cumulative distribution functions.

M = min(X1, X2, …, Xn)

\(P(min(X1, X2,..., Xn)>x) = P(X1>x, ..., Xn>x)\) $ = P(X1>x)…P(Xn>x)$ \(= e^{-\lambda_1 x} ... e^{-\lambda_n x}\) \(= e^{-(\lambda_1 + ... + \lambda_n)x}\)

The continuous distribution function of the minimun is an exponential with parameter \(\lambda_1 + ... + \lambda_n\)

The probability that the minimun is Xj is given by \(P(X_j < X_i \quad for\quad i\ne j)= \int_{0}^{\infty}{P(X_j<X_i for\quad i\ne j)}\lambda_je^{-\lambda_j x}dx\) \(= \int_{0}^{\infty}{\lambda_j e^{-\lambda_j}\prod_{i\ne j}{P(X_i>x)dx}}\) \(=\lambda_i \int_{0}^{\infty}{e^{-(\lambda_1 + ... + \lambda_n)x}dx}\) \(=\frac{\lambda_j}{\lambda_1 + ... + \lambda_n}\left[ -e^{-(\lambda_1 + ... + \lambda_n)x}\right]_{0}^{\infty}\)

\(=\frac{\lambda_j}{\lambda_1 + ... + \lambda_n}\) since \(\lambda_1= ...= \lambda_n =\lambda\), \(P(X_j < X_i \quad for\quad i\ne j)= \frac{1}{n}\) Then the mean of the minimun is \(\frac{\mu}n\)