Stochastic Processes Homework: Rewriting Expectation

[dirk_mec1]

Homework Statement

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'?

 

[statdad]

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.

 

[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?

 

[statdad]

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.

 

[dirk_mec1]

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?

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

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.

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?

 

[statdad]

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 $$f(x)$$?

 

[dirk_mec1]

What do you propose then?

 

[statdad]

To answer

What do you propose then?

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.

 

[dirk_mec1]

To answer


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.

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?

 

[statdad]

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.

 

[dirk_mec1]

When you write
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.

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?

 

[statdad]

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?

 

[dirk_mec1]

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

 

[statdad]

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.

 

[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!