# Expectation of random variable is constant?

1. Oct 7, 2006

### LM741

hi there.

currently looking at the two conditions that must be met for a process to be wide sense stationary.

The first constion is: E[X(t)] = constant

what exactly does this mean??isn't is obvious that any random variable (with fixed time) will always yield a constant expextation. I thought, for stationary prcesses, we want to try and prove that the random variable at DIFFERENT times yields the same expectation value (i.e. constant expactation).
The above condition seems to be stating the obvious...

Thanks
John

2. Oct 7, 2006

### steven187

Hi There,

Your sayin that, we are not in need of such a condition to satisfy the stationary concept, in actual fact we are in need of it especially when for example, lets says we are modeling the stock price historically it has been trading around $20 and then all of a sudden a stock split 1:4 occurs which then makes the stock trade at around$5 can you see the difference in the expectations before and after that particular event.

Regards Steven

3. Oct 7, 2006

### LM741

does your example incorporate fixed time??

also - can you tell me what the expectatino value of e^t is??
i.e. - E[ e^t] = ? not sure how to calculate this?

thanks steven

4. Oct 7, 2006

### LM741

just a follow up on my last post:

the reason why i'm not sure how to do this is because the expression does not contain a random variable , therefore how can i get a density function which i need in order to solve my expectation.
E[X] = integral(xfx)dx

where x is my random process and fx is the density function.

thanks

5. Oct 7, 2006

### steven187

Hi there,

of course if your looking at historical figure's then the time must be limited and therefore in a fixed time, in terms of your expectation it is suppose to be E[e^xt] this is the moment generating function which is an alternative to find the expectation to the integral x*f(x)dx. And as you can see
m(t)=E[e^xt]=integral e^xt*f(x) dx does involve the randam variable. To find the expectation E[x]=m'(0).

Regards

Steven