# Random walk:expectation

1. Jul 16, 2006

### Steve10

I have a random function f(n) which takes the values +/- 1 with equal probability.
Let the variable X take the sum of the values of f(n) after n steps. Then I can write,
$$X(T) = \sum_{n=0}^T f(n)$$ where T = 0,1,2,... and X(0) = 0.
And I can write the expectation of X as,
$$<X> = < \sum_{n=0}^T f(n) > = \sum_{n=0}^T <f(n) > = 0$$ since <f(n)> = 0 (by definition).
My question:
if, instead $$X(n) = \sum_{k=0}^n e^{-\mu k} f(n-k)$$ then, for the expectaion of X, can I write,
$$< X(n)> = \sum_{k=0}^n e^{-\mu k} < f(n-k) >$$
or even,
$$< X(n)> = \sum_{k=0}^n < e^{-\mu k}> < f(n-k) >$$
If I can do either of the above then can I also use <f(n)> = 0 to say that <X(n)> = 0, which I'm pretty sure it would have to be to answer the rest of my question.

TIA

Last edited: Jul 16, 2006
2. Jul 16, 2006

### mathman

You left out one important definition - what is mu? Is it a random variable or is it a constant? If it is random, how does it relate to f(n)? If it is constant or independent of f(n), then your conclusion is correct.

3. Jul 16, 2006

### Steve10

Thanks, I should have mentined that about mu. It is a constant value.
So I was correct after all.
I guess that's just as well, since I'm already typing up my results based on that conclusion!
Many thanks for the confirmation.

Last edited: Jul 16, 2006