1. ## Conditional Probabilities?

A high jumper achieves heights X0, X1... which are IID with pdf f and a strictly increasing F. We take X0 to be her first record. Her next record is achieved at "time T", where T=n if Xn>X0 and Xj<=X0, j=1,2...n-1. Show P(T=n) = 1/n(n+1)

P(T=n) = P(X0 is the highest among X0 to Xn-1 AND Xn is the highest among X0 to Xn)
Since they are independent,
= P(X0 is the highest among X0 to Xn-1)*P(Xn is the highest among X0 to Xn)
= 1/n * 1/(n+1)

because there is only 1/n chance for X0 to be the highest for the first X0 to Xn-1 (total n trials), and 1/n+1 chance for Xn to be the highest among the X0 to Xn (n+1 trials)

This is my thought, I just want to check if there is any problems if I approach the question in this way, because my TA keeps telling me I have to use conditional probabilities

2. Originally Posted by marko612
A high jumper achieves heights X0, X1... which are IID with pdf f and a strictly increasing F. We take X0 to be her first record. Her next record is achieved at "time T", where T=n if Xn>X0 and Xj<=X0, j=1,2...n-1. Show P(T=n) = 1/n(n+1)

P(T=n) = P(X0 is the highest among X0 to Xn-1 AND X0 is the highest among X0 to Xn)
Since they are independent,
= P(X0 is the highest among X0 to Xn-1)*P(X0 is the highest among X0 to Xn)
= 1/n * 1/(n+1)

because there is only 1/n chance for X0 to be the highest for the first X0 to Xn-1 (total n trials), and 1/n+1 chance for Xn to be the highest among the X0 to Xn (n+1 trials)

This is my thought, I just want to check if there is any problems if I approach the question in this way, because my TA keeps telling me I have to use conditional probabilities
I see two problems.

The first is a just typo:

P(T=n) = P(X0 is the highest among X0 to Xn-1 AND X0 is the highest among X0 to Xn)

should be

P(T=n) = P(X0 is the highest among X0 to Xn-1 AND Xn is the highest among X0 to Xn).

The second problem is with the logic. You have stated that the events "X0 is the highest among X0 to Xn-1" and "Xn is the highest among X0 to Xn" are independent, which means you can multiply the probabilities 1/n and 1/(n+1) to get the joint probability. But it is not obvious that those events are independent. You must prove that. From the definition of independence, that means you must prove

P(X0 is the highest among X0 to Xn-1 AND Xn is the highest among X0 to Xn)
= P(X0 is the highest among X0 to Xn-1)*P(Xn is the highest among X0 to Xn)

instead of just stating it. And the only way I know to do that is to start from the beginning and use conditional probabilities as your TA says. That is, you show the probability is the integral of a conditional probability:

$\displaystyle \int_{-\infty}^{\infty} f(x)F^{n-1}(x)(1-F(x))\ dx = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$

3. In my understanding, X0, X1..iid already implies those events are independent, so I am wrong??

4. Originally Posted by marko612
In my understanding, X0, X1..iid already implies those events are independent, so I am wrong??
It turns out those events are independent, but I and I think your TA are looking for a complete proof. Can you give one?

5. That's why I am a bit confused with this question. Based on the IID assumption, is it possible to say that they are independent without the proof? I know how to do the proof, but just wonder if this is really necessary...

6. Originally Posted by marko612
That's why I am a bit confused with this question. Based on the IID assumption, is it possible to say that they are independent without the proof? I know how to do the proof, but just wonder if this is really necessary...

It is necessary if you want to prove your assertion that the two events are independent. Call those events A and B. The definition of independence requires that you compute the joint probability of A and B and show P(A and B) = P(A)P(B). Can you do this without computing the conditional probabilities as I sketched? If so, please show me.

And you're welcome!

7. Originally Posted by marko612
That's why I am a bit confused with this question. Based on the IID assumption, is it possible to say that they are independent without the proof? I know how to do the proof, but just wonder if this is really necessary...

I've thought about this some more and I agree that there should be a simpler argument that the two events are independent. Here is my attempt at this. But it still uses conditional probabilities; I don't see any way around that.

Let A be the event X0 is the highest among X0 to Xn-1, and
let B be the event Xn is the highest among X0 to Xn.

Now P(A|B) = P(A) = 1/n since conditioning on event B will not change the relative probabilities that any Xi is the highest among X0 to Xn-1. Therefore

P(A and B) = P(A|B)P(B) = P(A)P(B),

so A and B are independent, and

P(T=n) = P(A and B) = P(A)P(B) = 1/n * 1/(n+1).

8. Originally Posted by JakeD
I've thought about this some more and I agree that there should be a simpler argument that the two events are independent. Here is my attempt at this. But it still uses conditional probabilities; I don't see any way around that.

Let A be the event X0 is the highest among X0 to Xn-1, and
let B be the event Xn is the highest among X0 to Xn.

Now P(A|B) = P(A) = 1/n since conditioning on event B will not change the relative probabilities that any Xi is the highest among X0 to Xn-1. Therefore

P(A and B) = P(A|B)P(B) = P(A)P(B),

so A and B are independent, and

P(T=n) = P(A and B) = P(A)P(B) = 1/n * 1/(n+1).

Thank you very much for your time, I better talk to the prof to see if he accepts this answer, because if there is a simplier way to solve the problem, I don't understand why we need to go through all the integration stuff to show they are independent. Thanks again.

9. Originally Posted by marko612
Thank you very much for your time, I better talk to the prof to see if he accepts this answer, because if there is a simplier way to solve the problem, I don't understand why we need to go through all the integration stuff to show they are independent. Thanks again.
Your question is a good one and interesting one. Please let us know what your prof says. Thank you.