God damn math

[Kuciwalker]

If X has the Poisson distribution with parameter L (lambda), show that:

E[X * (X - 1) * (X - 2) ... (X - k)] = L^(k+1), for k = 0, 1, 2...

I'm assuming all the X's are independent, in which case it's E[X] * E[X-1] ... E[X-k] = L * (L - 1) ... (L - k). Which doesn't equal L^(k+1).

If the X's aren't independent, it turns into a game of guess the value of the infinite sum. Ew.

[MRT144]

hows math helping you enjoy college?

[Kuciwalker]

By keeping me up until 2:30 AM doing problem sets.

[Kuciwalker]

And KH has apparently gone to bed. ****...

[Jon Miller]

Try using a Moment-Generating Function?

I think you will find this is true:

E(X^1)=L^1
E[X^2)=L^2+L

I guess this still has you trying to figure out sums.. (And I am not going to do that, it just might be easier to do from that side)

JM

[Eli]

I'm assuming all the X's are independent,

Eh? How on earth the Xs can be independent? They are the same and therefore as dependent as you can get.

[Eli]

The hard way to go about this is to open up the brackets and use the moment generating function to mop it up.

What JM said basically, but opening the brackets is annoying as hell. Well, I've done this **** two years ago, it's your turn now.

[Eli]

The easy way is induction.

[Lul Thyme]

Re: God damn math

Originally posted by Kuciwalker
If X has the Poisson distribution with parameter L (lambda), show that:

E[X * (X - 1) * (X - 2) ... (X - k)] = L^(k+1), for k = 0, 1, 2...

I'm assuming all the X's are independent, in which case it's E[X] * E[X-1] ... E[X-k] = L * (L - 1) ... (L - k). Which doesn't equal L^(k+1).

If the X's aren't independent, it turns into a game of guess the value of the infinite sum. Ew.

Right , because only KH knows math...
Why would the "X's" be independent, there is only one X....

I actually had to check my books for definitions because I havn't done anything close to this in a long time.
Like the others, my reflexes were induction, and then mfgs but it turns out its very direct:

E [X...(X-k)]
=Sum_{x=k+1} [x...(x-k)f(x)]
=Sum_{x=k+1} [x...(x-k)(L^x)e^(-L)/x!]
=L^(k+1) * e^(-L) * Sum_{x=k+1} [(L^x-k-1)/(x-k-1)!]
=L^(k+1) * e^(-L) * Sum_{x=0} [(L^x)/x!]
=L^(k+1) * e^(-L) * e^L
=L^(k+1)

note that the sum can clearly be started at k+1 since the terms will be 0 for x=0,...,x=k.
EDITED for clarity and technicalities.

[Lul Thyme]

Originally posted by Eli
The easy way is induction.

You would need to evaluate E[XY], where Y=(X-1)...(X-k) and E[Y]=L^k by the induction hypothesis.
I couldn't figure it out, and don't think it can be done except but doing the summation like I did.

[Lul Thyme]

Originally posted by Jon Miller
Try using a Moment-Generating Function?

I think you will find this is true:

E(X^1)=L^1
E[X^2)=L^2+L

I guess this still has you trying to figure out sums.. (And I am not going to do that, it just might be easier to do from that side)

JM

Yeah figuring out the resulting polynomial is problematic. The coefficients get nasty.

[Jon Miller]

Well, I was suggesting to come up with E(X^(k+1)) = L^(K+1)+Function(L,k), which would mean you wouldn't have to 'guess' what the sum is. This is just saying that the other direction might be easier.

JM
:lol, and I now am not sure what I was replying to. Either you DanSed me or I am getting sleepy.

[Lul Thyme]

Originally posted by Jon Miller
Well, I was suggesting to come up with E(X^(k+1)) = L^(K+1)+Function(L,k), which would mean you wouldn't have to 'guess' what the sum is. This is just saying that the other direction might be easier.

JM
:lol, and I now am not sure what I was replying to. Either you DanSed me or I am getting sleepy.

I did edit my quote.
I am not sure I understand what you are saying, let me reread it.

[Eli]

E[X(X-1)...(X-k)]
=E[X!/(X-k-1)!]
=Sum [x!f(x)/(x-k-1)!]
=Sum [x!(L^x)(e^-L)/x!(x-k-1)!]
=Sum [(L^x)(e^-L)/(x-k-1)!]
=(e^-L)L^(k+1) * Sum [(L^(x-k-1))/(x-k-1)!]
=(e^-L)L^(k+1) * (e^L)
=L^(k+1)

Yep. I was wrong.

I always had this problem of forgetting that you can do stuff the direct way like calculating E using the direct definition and not using wiseassry.

[Jon Miller]

You have edited your stuff like 3 times..

JM