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Let $X_{1}, . . . , X_{n}$ i.i.d. random variables with $X_{1} ∼ \text{Exp}(λ)$ for $λ > 0$. Show that $$\min_{1≤i≤n} X_{i} ∼ \text{Exp}(nλ)$$.

Mr.Gandalf Sauron
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aldd.
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  • This question is a direct copy of an exercise prompt without any reflection whatsoever. – Bcpicao Dec 07 '21 at 20:10

1 Answers1

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Hint:-

$P(\min(X_{i})\geq x)=P(X_{1}\geq x,X_{2}\geq x,....X_{n}\geq x)=\\ P(X_{1}\geq x)P(X_{2}\geq x)...P(X_{n}\geq x)=(P(X_{1}\geq x))^{n}=(1-(1-e^{-\lambda x}))^{n}=e^{-n\lambda x} $.

Now let $Y$ have $\text{Exp}(n\lambda)$ distribution. Then what is the expression for $P(Y\geq x)$?.

So what can you conclude from the fact that $P(Y\leq x) = 1- P(Y\geq x)$? Then what can you say about the CDF of Y and how does it compare with the cdf of $\min(X_{i})$ that we derived above?

Mr.Gandalf Sauron
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  • I thought about your hints and about your questions and I came to an conclusion but I am not sure if it is correct: P(Y≥x)= (P(Y≥x))^n = (1-(1-e^−nλx))^n=nλ e^−nλx – aldd. Dec 07 '21 at 21:32
  • Dude...I practically wrote the entire solution for you. you have $P(\min(X_{i})\geq x)= e^{-n\lambda x}\implies,P(\min(X_{i})\leq x) = 1-e^{-n\lambda x}=F(x)$ where $F$ is the cdf of a exponential $n\lambda$ variate. So the random variable $\min(X_{i})$ shares the same cdf of a exponential $n\lambda$ variate. Hence $\min(X_{i})\sim\exp(n\lambda)$. – Mr.Gandalf Sauron Dec 08 '21 at 06:27
  • If you are unwilling to show any work of your own I'll just delete my answer and vote to close your question. Not only are you violating the community guidelines , you are also not even trying to use mathjax. – Mr.Gandalf Sauron Dec 08 '21 at 06:37