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This problem appeared in a past Quiz. I'm still not able to solve it:

Note(λ = 1/B). Suppose X~exponential(λ = 1), Y ~ exp(λ = 1) and Z ~ exp ( λ = 3)

Find P(max(X,Y) > 10) and E(max(X,Z)).

I know that for exponential distribution

max(X,Y) = min(X,Y) + R. (Where R is the remaining lifetime).

However I don't know how to use this fact to solve a problem

Rad120
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1 Answers1

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\begin{align} & \frac d {dx} \Pr( \max\{X,Z\} \le x) = \frac d {dx} \Pr(X\le x\ \&\ Z\le x) \\[10pt] = {} & \frac d {dx} \big( (1 - e^{-x})(1-e^{-3x}) \big) = \frac d{dx}( 1 -e^{-x} - e^{-3x} + e^{-4x}) \\[8pt] = {} & e^{-x} + 3e^{-3x} - 4e^{-4x}. \end{align} So look for $\displaystyle\int_0^\infty x(e^{-x} + 3e^{-3x} - 4e^{-4x}) \, dx,$ remembering that $\displaystyle\int_0^\infty xe^{cx} \, dx = \frac 1 {c^2}. $

Michael Hardy
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  • $$ \begin{align} & \int_0^\infty x e^{cx},dx \{}\ = {} & \frac 1 {c^2} \int_0^\infty (cx) e^{-cx} (c,dx) \ {} \ = {} & \frac 1 {c^2} \int_0^\infty ue^{-u} , du. \end{align}$$ If that last integral is one you are very familiar with, then this substitution is the way to go. If you evaluate the last integral by parts, I'd write it as $$ \int u\big(e^{-u},du \big) = \int u,dv = \cdots $$ You can just as well use integration by parts directly on the integral you started with, but it's worth knowing what $$ \int u^n e^{-u},du $$ is, and recognizing the present problem as an instance. – Michael Hardy Nov 05 '20 at 19:39