Integral

How far have you got ?

what happens when you put in the limits ?

is it for an undergraduate course ?

are you sure the lower limit is not zero ?

In the notes we were given, E[h(x)] = Integral between plus and minus infinity (h(x)fx(x)dx)

So in this case h(x) = e^Zx and fx(x) = e^-x

Unless there is a reason to use 0 as the lower limit in this case, perhaps because an exponential rv is strictly positive?

i looked for it on a well-known search engine...

http://books.google.co.uk/books?id=BakFPFwrCfsC&pg=PA139&lpg=PA139&dq=%22moment+generating+function+of+an+exponentially+distributed+random+variable+%22&source=bl&ots=WzE8QxrGTS&sig=aIIxfZ8V3brQGmGz8WwUGY9tEw8&hl=en&sa=X&ei=EPjrUt3DAsixhAfJkAE&ved=0CCsQ6AEwAA#v=onepage&q=%22moment%20generating%20function%20of%20an%20exponentially%20distributed%20random%20variable%20%22&f=false

and they are starting at zero

In the notes we were given, E[h(x)] = Integral between plus and minus infinity (h(x)fx(x)dx)

So in this case h(x) = e^Zx and fx(x) = e^-x

Unless there is a reason to use 0 as the lower limit in this case, perhaps because an exponential rv is strictly non negative?

You're right, I assume that is because of the nature of an exponential rv, but now even using 0 instead of minus infinity, it still does not converge

Your integral as you've written it can't possibly converge because if Z-1 is positive it will diverge at one end of the integration range and if Z-1 is negative it will diverge at the other

You need to re-check your definition of the distribution or the valid values on which it is defined.

Problem solved, I missed the fact that z < 1 and therefore z-1 is negative.

The lower limit is infact 0 and not minus infinity as the exponential function is non negative.

Answer is 1/(1-Z)

Thanks to both of you!