# Stochastic processes

1. Sep 13, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data

I know that per definition $$E(N)= \sum P(N=k) \cdot k$$. But how can I rewrite the above expectation towards the 'usual definition'?

2. Sep 13, 2008

The expression

$$E(\mathbf{N}) = \sum_k \Pr(N=k) \cdot k$$

is the usual definition for the expectation of a discrete random variable. I'm not sure what alternative you refer to. You will see, in books on probability theory, this sum written in the form

$$E(\mathbf{N}) = \int x \, dP(x)$$

Edit: I was unable to view your attachment.

(a Riemann-Stieltjes integral), which reduces to the sum you (and I) have written. I doubt this is what you seek.

3. Sep 14, 2008

### dirk_mec1

I'm getting seriously tired of getting approval for the attachements. I'll postphone the first exercise for now.

But can someone help me with the second exercise (X is a stochast that attains non-negative values):

Proof the following:

$$E(X) = \int_0^{\infty} \overline{F}(x) \mbox{d}x$$

$$\overline{F}(x) = 1- F(x) = 1-P(X \leq x) = P(X>x)$$

What I've got so far

$$\int_{0}^{\infty} \overline{F}(x)\ \mbox{d}x = \int_0^{\infty} \int_x^{\infty} f(y)\ \mbox{d}y = \int_0^{\infty} \int_0^y f(x)\ \mbox{d}x = \int_0^{\infty} F(y)-F(0)\ \mbox{d}y = y \left( F(y) - F(0) \right)_{0}^{\infty} +E(y)$$

What's going wrong in my solution?

4. Sep 14, 2008

First, I've never seen the result you are trying to prove - but that doesn't prove it is written incorrectly. I know of the following result:

If the random positive random variable (so that $$F(0-) = 0$$, then

$$E(X) = \int_0^\infty \Pr(X > x) \, dx = \int_0^\infty \Pr(X \ge x) \, dx$$

although the integral may be infinite. Is this what you are discussing?

Second, your integrals, as written, don't make any sense - you need to integrate with respect to two variables, not just one. To point:

$$\int_0^\infty \, \int_x^\infty f(y) \, dy$$

is meaningless without a $$dx$$ as well.

5. Sep 14, 2008

### dirk_mec1

Well... its the expectation just written in another form which I'll have to proof.

Yes you're right so we get:

$$\int_{0}^{\infty} \overline{F}(x)\ \mbox{d}x = \int_0^{\infty} \int_x^{\infty} f(y)\ \mbox{d}y\ \mbox{d}x = \int_0^{\infty} \int_0^y f(x)\ \mbox{d}x\ \mbox{d}y = \int_0^{\infty} F(y)-F(0)\ \mbox{d}y = y \left( F(y) - F(0) \right)_{0}^{\infty} +E(y)$$

Statdad, what am I doing wrong here?

6. Sep 14, 2008

You are getting caught up in notation.

When you write (in the middle of your work)

$$\int_0^\infty \, \int_x^\infty f(y) \, dy dx = \int_0^\infty \, \int_0^y f(x) \,dx dy$$

why do you change the integrand from $$f(y)$$ to $$f(x)$$? You are correct in saying that the order of integration changes, so the inner integral is w.r.t. $$x$$, but do you really need to write $$f(x)$$?

7. Sep 15, 2008

### dirk_mec1

What do you propose then?

8. Sep 15, 2008

what you're doing in the double integral is not a change of variable - if that were the case, your step could be correct.
Try going through the same steps without changing from $$f(y)$$ to $$f(x)$$ at the aforementioned point. (And remember that when you integrate from $$0$$ to $$y$$ w.r.t. $$x, f(y)$$ will act like a constant.

9. Sep 15, 2008

### dirk_mec1

Thanks statdad It worked. Now returning to the first exercise:

I want to proof that (note that my sum runs to a finite value n)

$$E[N]= \sum_{k=1}^{n} P(N \geq k) = \sum k \cdot P(N=k)$$

So rewrite it to the usual definition.

Here's what I got so far (note the interchanging sum):

$$\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l) = \sum_{l=1}^n n \cdot P(N=l)$$

I'm stuck here! What going wrong here?

10. Sep 15, 2008

When you write

$$\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l) = \sum_{l=1}^n n \cdot P(N=l)$$

you are essentially writing

$$\Pr(N \ge k) = \sum_{l=1}^n \Pr(N=l)$$

This is not correct - the sum on the right here equals 1. In short, the expression for $$\Pr(N \ge k)$$ needs to be fixed.

Once that is one, you will have a double sum: reversing the order of summation (watch the indices) will get you where you need to be.

11. Sep 15, 2008

### dirk_mec1

Yes, you're right it should be:

$$\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=k}^n P(N=l) = \sum_{l=k}^n n \cdot P(N=l)$$

Is this correct?

12. Sep 15, 2008

This

$$\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l)$$

portion of your most recent post is still not correct (I haven't even included the final equality)

When you write out the sum for $$\Pr(N \ge k)$$, where should the summation begin - at $$1$$ or some other value?

13. Sep 15, 2008

### dirk_mec1

It should start at k but that's present in my post (#11) right?

14. Sep 15, 2008

I'm sorry, you are correct - I'm preparing for class and in my haste looked at the wrong post.

Yes, the first portion of your statement is correct.

$$\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=k}^n P(N=l)$$

Your error is in your next equality (which I did not include above)

Chew on this: The final double sum above can be written as

$$\sum_{\substack{k=1\dots,n\\l \ge k}} \Pr(N = l)$$

Similar to the other question dealing with double integration, think about how you can reverse the order of summation (inner sum over some values of $$k$$, outer sum of $$l$$ while maintaining the same inequality relationships between the two variables of summation.

15. Sep 16, 2008

### dirk_mec1

Thanks statdad I got it! You'll eventually get a summation that runs to 1 to l giving the l*P(N=l). I appreciate your effort in helping me!