Mixing
Ordering of maximums from exponential random variables
$begingroup$
Let $E_1,ldots,E_n$ be exponential random variables with parameters $(alpha_1,ldots,alpha_n)$. Further for $1leq zleq n$, let $M_1=max{E_1,ldots,E_z}$ and $M_2=max{E_{z+1},ldots, E_{n}}$. What is $mathrm{Pr}(M_1<M_2)$? If $z=n-1$ a simple expression emerges
$$
prod_{i=1}^{n-1}frac{alpha_i}{alpha_i+alpha_n}
$$
is there something as nice in the general case?
probability probability-theory probability-distributions
asked Jan 20 at 2:00
MichaelMichael
352213
$endgroup$
add a comment |
$begingroup$
Let $E_1,ldots,E_n$ be exponential random variables with parameters $(alpha_1,ldots,alpha_n)$. Further for $1leq zleq n$, let $M_1=max{E_1,ldots,E_z}$ and $M_2=max{E_{z+1},ldots, E_{n}}$. What is $mathrm{Pr}(M_1<M_2)$? If $z=n-1$ a simple expression emerges
$$
prod_{i=1}^{n-1}frac{alpha_i}{alpha_i+alpha_n}
$$
is there something as nice in the general case?
probability probability-theory probability-distributions
asked Jan 20 at 2:00
MichaelMichael
352213
$endgroup$
$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36
add a comment |
$begingroup$
Let $E_1,ldots,E_n$ be exponential random variables with parameters $(alpha_1,ldots,alpha_n)$. Further for $1leq zleq n$, let $M_1=max{E_1,ldots,E_z}$ and $M_2=max{E_{z+1},ldots, E_{n}}$. What is $mathrm{Pr}(M_1<M_2)$? If $z=n-1$ a simple expression emerges
$$
prod_{i=1}^{n-1}frac{alpha_i}{alpha_i+alpha_n}
$$
is there something as nice in the general case?
probability probability-theory probability-distributions
asked Jan 20 at 2:00
MichaelMichael
352213
$endgroup$
Let $E_1,ldots,E_n$ be exponential random variables with parameters $(alpha_1,ldots,alpha_n)$. Further for $1leq zleq n$, let $M_1=max{E_1,ldots,E_z}$ and $M_2=max{E_{z+1},ldots, E_{n}}$. What is $mathrm{Pr}(M_1<M_2)$? If $z=n-1$ a simple expression emerges
$$
prod_{i=1}^{n-1}frac{alpha_i}{alpha_i+alpha_n}
$$
is there something as nice in the general case?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 20 at 2:00
MichaelMichael
352213
asked Jan 20 at 2:00
MichaelMichael
352213
asked Jan 20 at 2:00
MichaelMichael
352213
asked Jan 20 at 2:00
MichaelMichael
352213
asked Jan 20 at 2:00
MichaelMichael
352213
352213
$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36
add a comment |
$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36
$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I can write it as a sum over permutations; if we use rate parameters $r_i=frac1{alpha_i}$, the probability that the exponentials are in order $E_{sigma(1)} < E_{sigma(2)} <cdots < E_{sigma(n)}$ is $frac{r_1r_2cdots r_n}{r_{sigma(n)}(r_{sigma(n)}+r_{sigma(n-1)})cdots(r_{sigma(n)}+r_{sigma(n-1)}+cdots+r_{sigma(1)})}$. In product and sum notation, that's
$$frac{prod_{i=1}^n r_i}{prod_{j=1}^nleft(sum_{k=j}^n r_{sigma(k)}right)}$$
Want the probability that some particular $E_i$ is the maximum? Sum over the permutations $sigma$ with $sigma(n)=i$. The probability that the maximum is in some subset $S$? Sum over $sigma$ with $sigma(n)in S$.
This is not pretty, and it gets more complicated as the number of exponentials grow, in stark contrast to the minimum that always stays nice. That minimum, by the way? If we sum over $sigma$ with $sigma(1)=i$, all of the terms that don't involve $r_{sigma(1)}=r_i$ balance out to $1$ and we get
$$frac{r_i}{sum_{j=1}^n r_j}$$
That doesn't happen with the maximum. I worked out $n=3$; the probability that $E_1$ is largest is
$$frac{r_2r_3(2r_1+r_2+r_3)}{(r_1+r_2)(r_1+r_3)(r_1+r_2+r_3)}=frac{alpha_1^2(alpha_1alpha_2+alpha_1alpha_3+2alpha_2alpha_3)}{(alpha_1+alpha_2)(alpha_1+alpha_3)(alpha_1alpha_2+alpha_1alpha_3+alpha_2alpha_3)}$$
For $n=4$, we're looking at sums of six terms. I could do that, but i'd need quite a bit more scratch space.
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
$endgroup$
add a comment |
$begingroup$
Let $F(x)=P(M_1le x)=prod_{i=1}^zP(E_ile x)$ and $G(x)=P(M_2le x)=prod_{i=z+1}^nP(E_ile x)$ $P(M_1le M_2|M_2=x)=F(x)$. By integrating over all values of $M_2$ we get $P(M_1le M_2)=int_0^infty F(x)G'(x)dx$.
For this special problem $P(E_ile x)=1-e^{-alpha_i x}$, writing it out looks messy.
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I can write it as a sum over permutations; if we use rate parameters $r_i=frac1{alpha_i}$, the probability that the exponentials are in order $E_{sigma(1)} < E_{sigma(2)} <cdots < E_{sigma(n)}$ is $frac{r_1r_2cdots r_n}{r_{sigma(n)}(r_{sigma(n)}+r_{sigma(n-1)})cdots(r_{sigma(n)}+r_{sigma(n-1)}+cdots+r_{sigma(1)})}$. In product and sum notation, that's
$$frac{prod_{i=1}^n r_i}{prod_{j=1}^nleft(sum_{k=j}^n r_{sigma(k)}right)}$$
Want the probability that some particular $E_i$ is the maximum? Sum over the permutations $sigma$ with $sigma(n)=i$. The probability that the maximum is in some subset $S$? Sum over $sigma$ with $sigma(n)in S$.
This is not pretty, and it gets more complicated as the number of exponentials grow, in stark contrast to the minimum that always stays nice. That minimum, by the way? If we sum over $sigma$ with $sigma(1)=i$, all of the terms that don't involve $r_{sigma(1)}=r_i$ balance out to $1$ and we get
$$frac{r_i}{sum_{j=1}^n r_j}$$
That doesn't happen with the maximum. I worked out $n=3$; the probability that $E_1$ is largest is
$$frac{r_2r_3(2r_1+r_2+r_3)}{(r_1+r_2)(r_1+r_3)(r_1+r_2+r_3)}=frac{alpha_1^2(alpha_1alpha_2+alpha_1alpha_3+2alpha_2alpha_3)}{(alpha_1+alpha_2)(alpha_1+alpha_3)(alpha_1alpha_2+alpha_1alpha_3+alpha_2alpha_3)}$$
For $n=4$, we're looking at sums of six terms. I could do that, but i'd need quite a bit more scratch space.
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
$endgroup$
add a comment |
$begingroup$
I can write it as a sum over permutations; if we use rate parameters $r_i=frac1{alpha_i}$, the probability that the exponentials are in order $E_{sigma(1)} < E_{sigma(2)} <cdots < E_{sigma(n)}$ is $frac{r_1r_2cdots r_n}{r_{sigma(n)}(r_{sigma(n)}+r_{sigma(n-1)})cdots(r_{sigma(n)}+r_{sigma(n-1)}+cdots+r_{sigma(1)})}$. In product and sum notation, that's
$$frac{prod_{i=1}^n r_i}{prod_{j=1}^nleft(sum_{k=j}^n r_{sigma(k)}right)}$$
Want the probability that some particular $E_i$ is the maximum? Sum over the permutations $sigma$ with $sigma(n)=i$. The probability that the maximum is in some subset $S$? Sum over $sigma$ with $sigma(n)in S$.
This is not pretty, and it gets more complicated as the number of exponentials grow, in stark contrast to the minimum that always stays nice. That minimum, by the way? If we sum over $sigma$ with $sigma(1)=i$, all of the terms that don't involve $r_{sigma(1)}=r_i$ balance out to $1$ and we get
$$frac{r_i}{sum_{j=1}^n r_j}$$
That doesn't happen with the maximum. I worked out $n=3$; the probability that $E_1$ is largest is
$$frac{r_2r_3(2r_1+r_2+r_3)}{(r_1+r_2)(r_1+r_3)(r_1+r_2+r_3)}=frac{alpha_1^2(alpha_1alpha_2+alpha_1alpha_3+2alpha_2alpha_3)}{(alpha_1+alpha_2)(alpha_1+alpha_3)(alpha_1alpha_2+alpha_1alpha_3+alpha_2alpha_3)}$$
For $n=4$, we're looking at sums of six terms. I could do that, but i'd need quite a bit more scratch space.
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
$endgroup$
add a comment |
$begingroup$
I can write it as a sum over permutations; if we use rate parameters $r_i=frac1{alpha_i}$, the probability that the exponentials are in order $E_{sigma(1)} < E_{sigma(2)} <cdots < E_{sigma(n)}$ is $frac{r_1r_2cdots r_n}{r_{sigma(n)}(r_{sigma(n)}+r_{sigma(n-1)})cdots(r_{sigma(n)}+r_{sigma(n-1)}+cdots+r_{sigma(1)})}$. In product and sum notation, that's
$$frac{prod_{i=1}^n r_i}{prod_{j=1}^nleft(sum_{k=j}^n r_{sigma(k)}right)}$$
Want the probability that some particular $E_i$ is the maximum? Sum over the permutations $sigma$ with $sigma(n)=i$. The probability that the maximum is in some subset $S$? Sum over $sigma$ with $sigma(n)in S$.
This is not pretty, and it gets more complicated as the number of exponentials grow, in stark contrast to the minimum that always stays nice. That minimum, by the way? If we sum over $sigma$ with $sigma(1)=i$, all of the terms that don't involve $r_{sigma(1)}=r_i$ balance out to $1$ and we get
$$frac{r_i}{sum_{j=1}^n r_j}$$
That doesn't happen with the maximum. I worked out $n=3$; the probability that $E_1$ is largest is
$$frac{r_2r_3(2r_1+r_2+r_3)}{(r_1+r_2)(r_1+r_3)(r_1+r_2+r_3)}=frac{alpha_1^2(alpha_1alpha_2+alpha_1alpha_3+2alpha_2alpha_3)}{(alpha_1+alpha_2)(alpha_1+alpha_3)(alpha_1alpha_2+alpha_1alpha_3+alpha_2alpha_3)}$$
For $n=4$, we're looking at sums of six terms. I could do that, but i'd need quite a bit more scratch space.
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
$endgroup$
I can write it as a sum over permutations; if we use rate parameters $r_i=frac1{alpha_i}$, the probability that the exponentials are in order $E_{sigma(1)} < E_{sigma(2)} <cdots < E_{sigma(n)}$ is $frac{r_1r_2cdots r_n}{r_{sigma(n)}(r_{sigma(n)}+r_{sigma(n-1)})cdots(r_{sigma(n)}+r_{sigma(n-1)}+cdots+r_{sigma(1)})}$. In product and sum notation, that's
$$frac{prod_{i=1}^n r_i}{prod_{j=1}^nleft(sum_{k=j}^n r_{sigma(k)}right)}$$
Want the probability that some particular $E_i$ is the maximum? Sum over the permutations $sigma$ with $sigma(n)=i$. The probability that the maximum is in some subset $S$? Sum over $sigma$ with $sigma(n)in S$.
This is not pretty, and it gets more complicated as the number of exponentials grow, in stark contrast to the minimum that always stays nice. That minimum, by the way? If we sum over $sigma$ with $sigma(1)=i$, all of the terms that don't involve $r_{sigma(1)}=r_i$ balance out to $1$ and we get
$$frac{r_i}{sum_{j=1}^n r_j}$$
That doesn't happen with the maximum. I worked out $n=3$; the probability that $E_1$ is largest is
$$frac{r_2r_3(2r_1+r_2+r_3)}{(r_1+r_2)(r_1+r_3)(r_1+r_2+r_3)}=frac{alpha_1^2(alpha_1alpha_2+alpha_1alpha_3+2alpha_2alpha_3)}{(alpha_1+alpha_2)(alpha_1+alpha_3)(alpha_1alpha_2+alpha_1alpha_3+alpha_2alpha_3)}$$
For $n=4$, we're looking at sums of six terms. I could do that, but i'd need quite a bit more scratch space.
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
answered Jan 20 at 3:08
jmerryjmerry
11.7k1527
11.7k1527
add a comment |
add a comment |
$begingroup$
Let $F(x)=P(M_1le x)=prod_{i=1}^zP(E_ile x)$ and $G(x)=P(M_2le x)=prod_{i=z+1}^nP(E_ile x)$ $P(M_1le M_2|M_2=x)=F(x)$. By integrating over all values of $M_2$ we get $P(M_1le M_2)=int_0^infty F(x)G'(x)dx$.
For this special problem $P(E_ile x)=1-e^{-alpha_i x}$, writing it out looks messy.
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
$endgroup$
add a comment |
$begingroup$
Let $F(x)=P(M_1le x)=prod_{i=1}^zP(E_ile x)$ and $G(x)=P(M_2le x)=prod_{i=z+1}^nP(E_ile x)$ $P(M_1le M_2|M_2=x)=F(x)$. By integrating over all values of $M_2$ we get $P(M_1le M_2)=int_0^infty F(x)G'(x)dx$.
For this special problem $P(E_ile x)=1-e^{-alpha_i x}$, writing it out looks messy.
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
$endgroup$
add a comment |
$begingroup$
Let $F(x)=P(M_1le x)=prod_{i=1}^zP(E_ile x)$ and $G(x)=P(M_2le x)=prod_{i=z+1}^nP(E_ile x)$ $P(M_1le M_2|M_2=x)=F(x)$. By integrating over all values of $M_2$ we get $P(M_1le M_2)=int_0^infty F(x)G'(x)dx$.
For this special problem $P(E_ile x)=1-e^{-alpha_i x}$, writing it out looks messy.
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
$endgroup$
Let $F(x)=P(M_1le x)=prod_{i=1}^zP(E_ile x)$ and $G(x)=P(M_2le x)=prod_{i=z+1}^nP(E_ile x)$ $P(M_1le M_2|M_2=x)=F(x)$. By integrating over all values of $M_2$ we get $P(M_1le M_2)=int_0^infty F(x)G'(x)dx$.
For this special problem $P(E_ile x)=1-e^{-alpha_i x}$, writing it out looks messy.
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
answered Jan 20 at 2:39
herb steinbergherb steinberg
2,8032310
2,8032310
add a comment |
add a comment |
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$begingroup$
That simple expression is obviously wrong; if the $alpha_i$ are all equal, the probability that $E_n$ is the greatest is $frac1n$, not $frac1{2^{n-1}}$. The probability that $E_n>E_i$ is $frac{alpha_n}{alpha_n+alpha_i}$, but the events aren't independent.
$endgroup$
– jmerry
Jan 20 at 2:09
$begingroup$
Dang that's an embarrassing mistake. Thanks for the answer, it was what I was hoping for.
$endgroup$
– Michael
Jan 20 at 19:36