1

Assume $A_i\sim\exp(a)$. Define $G=\max\{A_i\}_{i\in\{1,...,n\}}-\min\{A_i\}_{i\in\{1,\dots,n\}}$. Find $P(G\le x)$.

My step: $P(G\le x)$=$P(\max A_i\le\min A_i+x)=(\text{by Markov property})P(\max\ A_i\le \min\ A_i|\min\ A_i>x)=P(A_1=A_2=\dots A_n|\min A_i>x)=P(A_1=\dots=A_n)$ and $\min A_i>x)/P(\min\ A_i>x)$ I don't know which claims is wrong but there has to be something wrong cause i cannot get the answer

Thanks.

Davide Giraudo
  • 172,925
Jorden
  • 73

1 Answers1

1

The lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every $n\geqslant2$, $G$ is distributed as the maximum of $(n-1)$ i.i.d. random variables each exponentially distributed with parameter $a$. Since, however, the OP failed to explain their background, here is a direct, hands-on, approach.

Consider $U=\min\{A_k\,;\,1\leqslant k\leqslant n\}$ and $V=\max\{ A_k\,;\,1\leqslant k\leqslant n\}$, then $U\lt V$ almost surely and, for every $u\lt v$, $$ [u\lt U,V\lt v]=\bigcap_{k=1}^n[u\lt A_k\lt v], $$ hence, by independence of the random variables $(A_k)$, $$ P(u\lt U,V\lt v)=(\mathrm e^{-au}-\mathrm e^{-av})^n. $$ Differentiating this identity twice yields the density $f$ of $(U,V)$ as $$ f(u,v)=n(n-1)a^2\mathrm e^{-au-av}(\mathrm e^{-au}-\mathrm e^{-av})^{n-2}\mathbf 1_{0\lt u\lt v}. $$ By definition, $G=V-U$ hence, for every $x\gt0$, $P(G\leqslant x)=(\ast)$ with $$ (\ast)=\int_0^\infty\!\!\!\int_u^{u+x}f(u,v)\mathrm dv\mathrm du=\int_0^\infty na\mathrm e^{-au}\left[(\mathrm e^{-au}-\mathrm e^{-av})^{n-1}\right]_{v=u}^{v=u+x}\mathrm du, $$ that is, $$ (\ast)=\int_0^\infty na\mathrm e^{-au}(\mathrm e^{-au}-\mathrm e^{-au-ax})^{n-1}\mathrm du=(1-\mathrm e^{-ax})^{n-1}\int_0^\infty na\mathrm e^{-nau}\mathrm du, $$ and finally, $$ P(G\leqslant x)=(1-\mathrm e^{-ax})^{n-1}. $$ To the OP: In the question you assert that $P(V\leqslant U+x)=P(V\leqslant U\mid U\leqslant x)$ and you explain that this holds "by Markov property". This is obviously wrong but the trouble is that I cannot even see what you think you are doing there... Please explain.

Did
  • 279,727
  • 1
    I think this answer is very clear but one statement, " lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every $n\ge 2$, G is distributed as the maximum of (n−1) i.i.d. random variables each exponentially distributed with parameter a." I wonder how can you deduce from markov property? – Jorden Mar 14 '14 at 04:54
  • Did you read my comment to the question? Please do. The statement you quote does not even mention the name "Markov". – Did Mar 14 '14 at 05:34
  • Isn't markov property the same things as the lack of memory property? – Jorden Mar 14 '14 at 05:36
  • No it is not. Please see added paragraph at the end of my answer. – Did Mar 14 '14 at 05:41
  • so can you explain a bit how it is related to lack of memory property? i dont quite see why you know it is the distribution of n-1 random variable from this property. – Jorden Mar 14 '14 at 05:42
  • This is becoming a little tiresome... No there is no "distribution of n-1 random variable(s)" in my answer (so what are you talking about?) and unless you explain precisely why you think that Markov property and lack of memory should be "related" (which you don't at the moment), what can I say? The things you seem to believe are most certainly at least partly wrong but you have to be much more specific if you want somebody to debunk them. – Did Mar 14 '14 at 05:48
  • I treat the markov property the same as lack of memory property. We know that the $minX_i$ is a exponential random variable so by the lack of memory property, $P(max\le min|min>x)=P(max-x\le min)=P(max\ge min +x)$ – Jorden Mar 14 '14 at 05:50
  • 1
    Pure nonsense, sorry. The fact that the minimum is exponential does not imply this, in any way. – Did Mar 14 '14 at 05:52
  • not all exponential random variable has lack of memory property? And i made some mistake about the inequality sign – Jorden Mar 14 '14 at 05:53
  • Define lack of memory, which involves one random variable, and compare to what you write, which involves two random variables. – Did Mar 14 '14 at 05:55
  • ya, i know hwat you mean, but a random variable is some real value, but whatever value it is, as long as we know the $min>x$ the distribution of $max\le min | min >x$and $max-x\le min$ should be the same. But i guess i should have something wrong as you seem an expert in it. I would ponder it. – Jorden Mar 14 '14 at 06:02
  • No. (Any source for what you write?) – Did Mar 14 '14 at 06:03