question on min(x_1,x_2) (1 Viewer)

[pLuvia]

How do you interpret the min(x₁,x₂) for example

how do you solve this

Gx₁+x₂(z)=sigma(x₁=1 to infinity) sigma(x₂=1 to infinity) z^{min(x₁,x₂)} f(x₁)f(x₂)

I'm not entirely sure how interpret the min(x_1,x_2) bit

 

[darkliight]

x_1 if x_1 < x_2, x_2 otherwise?

 

[pLuvia]

Oh I see thanks

Do you by any chance know how to solve that thing, it should come out to be a "nice" expression (hopefully). I'm not sure how to do the summation of the z^min(x_1,x_2)

 

[darkliight]

No idea sorry

 

[Slidey]

That looks very abstract pLuvia. Are you sure you can 'simplify' that?

Dunno if this helps: http://functions.wolfram.com/ElementaryFunctions/Min/02/
And: http://functions.wolfram.com/ElementaryFunctions/Min/

 

[pLuvia]

Oh crap, forgot to add what the function was my bad. Sorry

Umm, f(x)=(1-p)^x-1p where x=1,2,3...

This is to do with probability generating functions. I've tried doing an incremental case with just the x_1, x_2 values going up to 3, but it didn't really turn out nice enough to produce a nice enough expression

Oh and thanks for those sites slidey

So the whole question is to find a nice expression for this

Where f(x)=(1-p)^x-1p where x=1,2,3...

Gx₁+x₂(z)=sigma(x₁=1 to infinity) sigma(x₂=1 to infinity) z^{min(x₁,x₂)} f(x₁)f(x₂)

 

[pLuvia]

Hmm, I'll look into that. Thanks

 

[darkliight]

z^min(x1,x2) means you have to consider two cases x1 < x2 or x2 < x1, both cases are the same because of the 2 summations.

Umm, the summation modifies the values of x_1 and x_2? In other words, you can't consider two cases separately. x_1 will not always be < x_2, or vice versa.

 

[Slidey]

Umm, the summation modifies the values of x_1 and x_2? In other words, you can't consider two cases separately. x_1 will not always be < x_2, or vice versa.

That's what I'm thinking, because the values of x1 and x2 depend on the current values in the summations. Good to see 3unitz have a crack. Personally have no clue how to begin to approach it.

 

[Affinity]

Oh crap, forgot to add what the function was my bad. Sorry

Umm, f(x)=(1-p)^x-1p where x=1,2,3...

This is to do with probability generating functions. I've tried doing an incremental case with just the x_1, x_2 values going up to 3, but it didn't really turn out nice enough to produce a nice enough expression

Oh and thanks for those sites slidey

So the whole question is to find a nice expression for this

Where f(x)=(1-p)^x-1p where x=1,2,3...

Gx₁+x₂(z)=sigma(x₁=1 to infinity) sigma(x₂=1 to infinity) z^{min(x₁,x₂)} f(x₁)f(x₂)

Oh this makes it different.. and easier. I knew it was something related..

If you consider 2 random variables X1 and X2 with geometric distribution given by the fs.. (that is with parameter p) then Y=min(X1,X2) is another geometric random variable with parameter q=[1-(1-p)^2]. (think why!)
so G(z)=E(z^Y) is just the probability generating function for geometric([1-(1-p)^2])

so G(z) = zq/(1-(1-q)z) From memory. (it's just a simple geometric series if you want to do it bare hands)

in terms of p, it is z(1-(1-p)^2)/(1-z(1-p)^2)

compare with the continuous analog with exponential random variables.
if you have exponential(a_1), exponential(a_2) .. so on
their minimum has distribution exponential(a1+a2+...)

 

[Affinity]

made a mistake when i summed z^x (1-p)^(x-1)

should be z / (1 - z + zp) not z/(1-z) . 1/p

so should be Gx1 + x2(z) = zp / (1 - z + zp)

:/

I think this is wrong,

zp/(1-z+zp) = \sum_{n=1}^\infty z^n f(n) though

 

[Affinity]

z^min(x1,x2) means you have to consider two cases x1 < x2 or x2 < x1, both cases are the same because of the 2 summations.

for, x1 < x2
Sigma(x1=1 to infinity) Sigma(x2=1 to infinity) z^x1 p(1 - p)^(x1-1) . p(1 - p)^(x2-1)
= p^2 Sigma(x1=1 to infinity) z^x1 (1 - p)^(x1-1) Sigma(x2=1 to infinity) (1 - p)^(x2-1)
= p^2 [z/(1 - z) . 1/p] [1/p]
= z/(1 - z)

for, x2 < x1
Sigma(x1=1 to infinity) Sigma(x2=1 to infinity) z^x2 p(1 - p)^(x1-1) . p(1 - p)^(x2-1)
= p^2 Sigma(x1=1 to infinity) (1 - p)^(x1-1) Sigma(x2=1 to infinity) z^x2 (1 - p)^(x2-1)
= p^2 [1/p] [z/(1 - z) . 1/p]
= z/(1 - z)

Gx1 + x2(z) = z / (1 - z)

sorry if its wrong i have no idea what this is :/

you can't sum like that. If you condition, then you would need to write

for, x1 <= x2
Sigma(x1=1 to infinity) Sigma(x2=x_1 to infinity) z^x1 p(1 - p)^(x1-1) . p(1 - p)^(x2-1)
= ...

for, x2 < x1
Sigma(x1=1 to infinity) Sigma(x2=1 to x_1-1) z^x2 p(1 - p)^(x1-1) . p(1 - p)^(x2-1)
= ...

 

[pLuvia]

Oh this makes it different.. and easier. I knew it was something related..

If you consider 2 random variables X1 and X2 with geometric distribution given by the fs.. (that is with parameter p) then Y=min(X1,X2) is another geometric random variable with parameter q=[1-(1-p)^2]. (think why!)
so G(z)=E(z^Y) is just the probability generating function for geometric([1-(1-p)^2])

so G(z) = zq/(1-(1-q)z) From memory. (it's just a simple geometric series if you want to do it bare hands)

in terms of p, it is z(1-(1-p)^2)/(1-z(1-p)^2)

compare with the continuous analog with exponential random variables.
if you have exponential(a_1), exponential(a_2) .. so on
their minimum has distribution exponential(a1+a2+...)

I'll ponder on that one, thanks

 

[pLuvia]

Would you be able to give me a slight hint on finding the geometric parameter for the transformation Y=min(x_1,x_2), has it anything to do with the geometric cdf? Because the cdf looks like the transformed parameter when the power of (1-p) is 2

 

[Affinity]

You can go along the lines of:
P(min(x1,x2)>n) = P(x1>n and x2 > n) = P(x1 > n)P(x2>n) = [(1-p)^2]^n


or you can argue that a geometric random variable is the number of indepedent trials needed to record a success.. now with min(x1,x2) is the smaller of these so you can look at it as 1 geometric rv, with a success whenever x1 or x2's trials is a success.

 

[pLuvia]

I'm sorry I really don't understand why you made the min greater than n and you got out the last expression. I haven't really done any order statistics to tell you the truth.

Also could you explain the other method you stated

thanks

 

[Affinity]

that results in a simpler expression than using P(* <=n) since P(A and B) = P(A)P(B) but you can;t say the same with P(A or B) (when A, B are independent)

with the other one you will have to interpret the meaning of "a geometric rv"