probability generating functions Watch

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kikzen
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hopefully someone can help me out with this

Suppose the PGF of a RV Z is given by:

G_z(S) = (4-S) / (3)(4-3s)

Find the distribution of Z, that is, the probabilities
P(Z = i) for i = 0, 1, 2, . . . .

my tutor said something like get it into the form a + 1/(1-s) but i cant... help !
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Gauss
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(Original post by kikzen)
hopefully someone can help me out with this

Suppose the PGF of a RV Z is given by:

G_z(S) = (4-S) / (3)(4-3s)

Find the distribution of Z, that is, the probabilities
P(Z = i) for i = 0, 1, 2, . . . .

my tutor said something like get it into the form a + 1/(1-s) but i cant... help !
The Probability Generating Function (hereafter G_z(s)) is defined by the following:

G_z(S) = E_z(s^z) = Sum_[ i=0..infinity] s^(z_i)*f_z(z_i)

The probability f_z(z) = P(Z=z) is the coefficient of the zth power of S. Equate it all to (4-S) / (3)(4-3s) and get it in the form a + 1/(1-s) and it will be clear what it should look like from there.

If you still can't see it let me know.

Euclid.
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kikzen
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yah i know all that fool :p:

it was just the rearranging but ive figured out the problem!
i was reading s/6 as 5/6. ah the joys of stupidity.

thanks anyway !
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Gauss
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(Original post by kikzen)
yah i know all that fool :p:

it was just the rearranging but ive figured out the problem!
i was reading s/6 as 5/6. ah the joys of stupidity.

thanks anyway !
Lol sometimes it gets so bad that I cannot even read my own writing!

Euclid.
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kikzen
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Let p E (0, 1) and suppose that a RV X has
P(X = i) = p(1 − p)^i (i = 0, 1, 2, . . . )

Find the PGF for X, G_x(s) := E(s^x) for s E [0, 1].

Right well i wrote out the sum and that was a powerseries so G_x(s) =
p/(1-s(1-p))



Suppose that Y is independent but identically distributed to X.
so that means G_x(s) = G_y(s)

Deduce the PGF, G_(x+y) (s), for the sum X +Y .
well theyre equal so it is
[p/(1-s(1-p))]^2


Use the PGF to find the
mean of X +Y . By expanding GX+Y (s) as a power series in s, find the
distribution of X + Y .

this is where i get stuck. i know its another rearrangement but icant see how to do it at all, with the squareds everywhere!

help! (i may have done something wrong before so point that out and i might be able to figure it out ) danke!
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Gauss
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(Original post by kikzen)
Suppose that Y is independent but identically distributed to X.
so that means G_x(s) = G_y(s)
Are you sure they're equal? As Y is iid to X, then shouldn't G_y(s) = [G_x(s)]^n?

Euclid.
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kikzen
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why?
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Gauss
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(Original post by kikzen)
why?
Sorry I should have said G_y(s) = [G_x(s)]^(1/n)

Here's why.

Euclid.
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