probability question

[YoungGun21]

I have the sum from k=1 to infinity of P(X >= k). What are some possible ways I can manipulate this? X is a discrete r.v.

I did it and got P(X=j)*P(X=k), where j=k+1 and j goes from 0 to infinity, k still going from 1 to infinity. This doesn't seem to be right though because after switching the order of summation I just get that this whole thing equals 1, when I think it should equal k..?

Any help?

 

[busydude]

P(x>=k) = 1 - p(x<k)

 

[YoungGun21]

P(x>=k) = 1 - p(x<k)

How does this change my bounds of summation though? Or do they not change at all?

 

[busydude]

Are you sure there is a sum of k from 0 -> inf in the problem? Or.. did they just specify the range of K?

 

[YoungGun21]

Are you sure there is a sum of k from 0 -> inf in the problem? Or.. did they just specify the range of K?

Yes I'm sure. It is k from 1 to infinity. I'm actually trying to show that this sum (P(X>=k))is the same as the sum k=0 to infinity of P(X>k)

 

[DrPizza]

How does this change my bounds of summation though? Or do they not change at all?

Oh, that's all you're trying to do? If you let j= k-1, then when k=1, j would = 0. So the lower limit of summation would be j=0, and since j=k-1, you should be able to figure out what to substitute for k.

 

[busydude]

I'm actually trying to show that this sum (P(X>=k))is the same as the sum k=0 to infinity of P(X>k)

Oh.. I see, brb in a moment.

 

[busydude]

Oh, that's all you're trying to do? If you let j= k-1, then when k=1, j would = 0. So the lower limit of summation would be j=0, and since j=k-1, you should be able to figure out what to substitute for k.

Yup, this is your answer.

 

[YoungGun21]

Alright thanks for your help. I think there must be some summation tricks I'm missing since I'm still not getting the answer I need.

 

[busydude]

It is pretty simple...

Summation of ...... P(X>=K) = Summation of ...... P(X=1)+P(X=2)+.........+P(X=K) ----- (1)
Summation of ...... P(X>K) = Summation of ....P(X=0)+P(X=2)+.........+P(X=K-1) ------ (2)

Just substitute j=k+1 in equation (1) and then you can easily prove that eq (1) = eq(2)