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

Short Answer

Expert verified

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

Step by step solution

01

Given information

The random variable.\({X_1},{X_2},..{X_k}\) are independent. \({X_i}\) has the exponential distribution.

02

Computing the probability

For any number\(y > 0\),

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

\(\begin{array}{c}{\rm P}\left( {Y > y} \right) = \exp \left( {{\beta _1}y} \right)...\exp \left( { - {\beta _k}y} \right)\\ = \exp \left( { - \left( {{\beta _1} + ... + {\beta _k}} \right)y} \right)\end{array}\)

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

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Most of the calculation in Example 3.8.4 is quite general.Suppose that X has a continuous distribution withp.d.f. \({\bf{f}}\) . Let\({\bf{Y = }}{{\bf{X}}^{\bf{2}}}\), and show that the p.d.f. of Y is

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Letnandkbe positive integers such that bothnandn−kare large. Use Stirling’s formula to write as simple an approximation as you can forPn,k.

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