Solve E(min(X,100)) Using Geometric Distribution with Theta | Homework Help

[sneaky666]

Homework Statement

Whats a hint to solve E(min(X,100)), when X~Geometric(theta)?

Homework Equations

geometric distribution where p is theta

The Attempt at a Solution

I got here

99
summation x*theta*(1-theta)^x
x=o

+

inf.
summation 100*theta*(1-theta)^100
x=100

But I don't know what to do now.

 

[statdad]

Two suggestions:
1) Don't write your summations using x as the index of summation: if you wanted to compute E(X) you could write it as

<br /> \sum_{j=0}^\infty j \theta (1-\theta)^j<br />

Do something similar in your problem.

2) Informally, if X = 100 then it happens that \min(X, 100) = X,
so your first sum can extend to 100 and the second can start at 101.

Thus your expression can be written as

<br /> E[\min(X,100)] = \sum_{j=0}^{100} {j \, \theta (1-\theta)^j} + \sum_{j=101}^\infty {100 \, \theta (1-\theta)^j}<br />

What can you do with this?

 

[sneaky666]

for your second summation, why is it to the power of j, shouldn't it be to the power of 100, since its a constant probability when x>= 100?

 

[statdad]

for your second summation, why is it to the power of j, shouldn't it be to the power of 100, since its a constant probability when x>= 100?

No - that's part of the reason I suggested changing the index of summation away from X. You use the distribution of X to calculate the expected value - the exponent in that distribution is not a constant. Perhaps it would help to think about the second sum as involving the expectation of a constant function with respect to the distribution of X.

 

[sneaky666]

ok, but what is the next step, how do i expand the summations?

 

[statdad]

Work on them and come back with what you tried.

 

[sneaky666]

so then the second summation can be changed to

theta(1-theta) * summation from k=100 to infinity of k(1-theta)^(k-1)
theta(1-theta) * summation from k=100 to infinity of (-d/d*theta)*(1-theta)^k
theta(1-theta) *(-d/d*theta)* summation from k=100 to infinity of (1-theta)^k
theta(1-theta) *(-d/d*theta)((1-theta)^100)/theta)
theta(1-theta) * (100theta(1-theta)^99 + (1-theta)^100)/theta^2
(1-theta) * (100theta(1-theta)^99 + (1-theta)^100)/theta
((1-theta)/theta) - (1-theta) * (100theta(1-theta)^99 + (1-theta)^100)/theta
this new term in the begginning is the expected value of a geometric distribution
((1-theta)(1-(1-theta)^100))/theta - (100theta(1-theta)^100) / theta
here i am stuck, i know this is not the answer but it is pretty close

so my last step is
99
summation ktheta(1-theta)^k
k=0
+
((1-theta)(1-(1-theta)^100))/theta
-
100(1-theta)^100

the answer i think is the middle term, but somehow i need to remove the first and third term...