Hello all, I have a iid exponential random variables question with a few parts and would like guidance on approaching it. Not sure if I'm on the right track, and any help would be greatly appreciated.
Problem statement:
Attempt/thoughts
Since W1, W2, ... Wn are continuous RVs, there is zero probability that they have the same value. So, we have Wa > Wb > Wc > ..., where (a, b, c, ...) is a permutation of (1, 2, 3, ...) and since the variables are i.i.d., each of these permutations would be equally likely. So we have:
P(W1 > W2) 1/2
P(W1 > W2 > W3) = 1/6
P(W1 > W2 > W3 > ... > Wn) = 1/n!
(a) P(W1 > W2) =1- P(W2 < W1) = 1/2
(b) Thinking of this problem as 3 numbers independently from Exp(1) getting W1,..,W3. Writing the results in ascending order, by symmetry, since you draw independently from the same distribution, any permutation of (W1,W2,W3) will be equally likely to occur. We are told that W2 > W1 in the conditional probability. So, we are left with only 3 possible permutations:
W1 < W2 < W3
W2 < W1 < W3
W3 < W2 < W1
Each outcome is equally likely and so, P(W3 > W2 | W2 > W1) = 1/3
(c) Same reasoning as above. Thinking of this problem as 10 numbers independently from Exp(1) getting W1,..,W10. Writing the results in ascending order, by symmetry, since you draw independently from the same distribution, any permutation of (W1,W2,...,W10) will be equally likely to occur. We are told that W9 > W8 > ... > W1 in the conditional probability. So, we are left with only 10 possible permutations:
W10<W1<...<W9
W1<W10<...<W9
W1<W2<W10<...<W9
...
W1<...<W10<W9
W1<...<W9<W10
Each outcome is equally likely and only the last one fulfills W10>W9. So, P(W10>W9 | W1<...<W9) = 1/10
Am I completely off track here or should I be finding the pdfs and integrating?
Problem statement:
Attempt/thoughts
Since W1, W2, ... Wn are continuous RVs, there is zero probability that they have the same value. So, we have Wa > Wb > Wc > ..., where (a, b, c, ...) is a permutation of (1, 2, 3, ...) and since the variables are i.i.d., each of these permutations would be equally likely. So we have:
P(W1 > W2) 1/2
P(W1 > W2 > W3) = 1/6
P(W1 > W2 > W3 > ... > Wn) = 1/n!
(a) P(W1 > W2) =1- P(W2 < W1) = 1/2
(b) Thinking of this problem as 3 numbers independently from Exp(1) getting W1,..,W3. Writing the results in ascending order, by symmetry, since you draw independently from the same distribution, any permutation of (W1,W2,W3) will be equally likely to occur. We are told that W2 > W1 in the conditional probability. So, we are left with only 3 possible permutations:
W1 < W2 < W3
W2 < W1 < W3
W3 < W2 < W1
Each outcome is equally likely and so, P(W3 > W2 | W2 > W1) = 1/3
(c) Same reasoning as above. Thinking of this problem as 10 numbers independently from Exp(1) getting W1,..,W10. Writing the results in ascending order, by symmetry, since you draw independently from the same distribution, any permutation of (W1,W2,...,W10) will be equally likely to occur. We are told that W9 > W8 > ... > W1 in the conditional probability. So, we are left with only 10 possible permutations:
W10<W1<...<W9
W1<W10<...<W9
W1<W2<W10<...<W9
...
W1<...<W10<W9
W1<...<W9<W10
Each outcome is equally likely and only the last one fulfills W10>W9. So, P(W10>W9 | W1<...<W9) = 1/10
Am I completely off track here or should I be finding the pdfs and integrating?
