Answer:
Now we can find the expected value with this formula:
And replacing we got:
Step-by-step explanation:
For this case we know the following probability mass function given:
For this case we need to satisfy the following condition in order to have a probability distribution function:
And we have this:
Now we can find the expected value with this formula:
And replacing we got:
If you know that
then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving .
For starters, we have
Let . Then as
, we also have
, and
So in terms of , the limit is equivalent to
Now use some of the properties of limits: the above is the same as
The first limit is trivial; , so its value is 1. The second limit comes out to
To see why this is the case, replace , so that
as
, and
Then the limit we're talking about has a value of
# # #
Another way to do this without knowing the definition of as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write
(where the notation means , just to get everything on one line).
Recall that
if is continuous at
.
is continuous everywhere, so we have
For the remaining limit, write
Now as , both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to
and our original limit comes out to the same value as before, .