7.2, 10:

CDF:

\(F_{X_i}(x) = P(X_i \leq x)= 1-\lambda e^{-\lambda x}\)

\(F(X)=1-\lambda e^{-\lambda x}\). Let \(M\) be \(M=\min\{X_1, X_2, ...,X_n\}\). The cummulative distribution of \(M\) is then given by:

\(F_M(y) = P(M \leq y)=1-P(M \geq y)=1-P(\min\{X_1, X_2, ...,X_n\} \ge y)\) \(=1-P(X_1 \leq y,X_2 \leq y,...,X_n \geq y)\) \(=1-P(X_1 \ge y)\cdot(P(X_2 \le y) \cdot \cdot \cdot(P(X_n \le y)\) \(=1-e^{-\lambda_1y}\cdot e^{-\lambda_2 y}\cdot \cdot \cdot e^{-\lambda_n y}\) \(=1-e^{-\lambda_1y-\lambda_2 y-\cdot \cdot \cdot\lambda_n y}\) \(=1-e^{-\lambda_1y-\lambda_2 y-\cdot \cdot \cdot -\lambda_n y}\) \(=1-e^{-\sum_{i=1}^n{\lambda_i y}}\)

Since the density function is the first derivative of the CDF:

\(\frac{d}{dy}F_{M} (y)=\sum_{i=1}^n \lambda_i e^{-\lambda y}\)

Since the expected value of a single exponential distribution is defined as:

\(E[X]=\frac{1}{\lambda}\). This problems provides that each \(X_i\) has mean \(\mu\). Then this leads to \(E(M)=\frac{\mu}{\lambda}\)

References: Exponential Distro and this theorem.