Question

Suppose that the random variables\({X_1},...,{X_k}\)are independent and\({X_i}\)has the exponential distribution with parameter\({\beta _i}\left( {i = 1,...,n} \right)\). Let\(Y = \min \left\{ {{X_{1,...,}}{X_k}} \right\}\)Show that Y has the exponential distribution with parameter\({\beta _1} + .... + {\beta _k}\).

Short answer

It is proved that Y has the exponential distribution with parameter \({\beta _1} + {\beta _2} + ... + {\beta _k}\).

Step-by-step solution

Given information

A random variables \({X_1},...,{X_k}\), are independent and \({X_i}\) follows exponential distribution with parameter\({\beta _i}\).

Showing that Y has the exponential distribution.

Let\(Y = \min \left\{ {{X_{1,...,}}{X_k}} \right\}\)

For any number \(y > 0\)

\(\begin{aligned}{}{\rm P}\left( {Y > y} \right)& = {\rm P}\left( {{X_1} > y,...,{X_k} > y} \right)\\& = {\rm P}\left( {{X_1} > y} \right)...{\rm P}\left( {{X_k} > y} \right)\end{aligned}\)

Since, by definition of exponential distribution,

\(P\left( {X > x} \right) = {e^{ - \beta x}}\)

Therefore,

\(\begin{aligned}{}{\rm P}\left( {Y > y} \right)& = {e^{ - {\beta _1}y}} \cdot {e^{ - {\beta _2}y}} \cdots {e^{ - {\beta _k}y}}\\ &= {e^{ - \left( {{\beta _1} + ... + {\beta _k}} \right)y}}\end{aligned}\)................................(1)

Equation (1) gives the probability that an exponential random variable with parameter \({\beta _1} + ... + {\beta _k}\) is greater than y. Hence Y has that exponential distribution with parameter\({\beta _1} + ... + {\beta _k}\).