MHB Is this algebraic expression true for q+p=1?

[Poirot1]

show that $\frac{(\frac{q}{p})^{k-a}-(\frac{q}{p})^k}{1-(\frac{q}{p})^k}=\frac{1-(\frac{q}{p})^a}{1-(\frac{q}{p})^k}$
where q+p=1

Thanks

 

[Wilmer]

Since denominators are the same, WHY do you bother showing them?
(if a/x = b/x, then a = b)

 

[Poirot1]

you are right but that is irrevelant, I have tried to show numerators are equal and have failed.

 

[Wilmer]

you are right but that is irrevelant, I have tried to show numerators are equal and have failed.

Looks to me like your whole problem is irrelevant !

Assign values, say q=2, p=-1, k=2, a=1
left numerator = -6, right numerator = 3
So STOP!

 

[Poirot1]

I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?

 

[Wilmer]

I perhaps should have mentioned that p and q are probabilities, so must be between 0 and 1. Does that make a difference?

Equal (both = 0) if p = q = 1/2
Also equal if a = k, of course.

Sir Poirot, just noticed I received this compliment from you:
"Have I not stated that p+q=1, which you, in your eagerness to be contemptous, have ignored."
May I humbly defend myself by reminding you thay my post says: q=2 and p=-1 : 2 + (-1) = 1.

 

[HallsofIvy]

If q= 2/3, p= 1/3, so that p and q are both between 0 and 2 and add to 1, the numerator becomes 2^{k-a}- 2^k= 1- 2^a. And, if k=2, a= 1, that says 2- 4= 2 which is certainly not true. You cannot prove what you want, it is not true.

 

[Mathwebster]

Just an added note. If you factor the numerator then you have

$\left(\dfrac{q}{p}\right)^{k-a}\left(1 - \left(\dfrac{q}{p}\right)^a\right)$

Comparing with the right hand side give that

$\left(\dfrac{q}{p}\right)^{k-a}=1$ for your equality to be true. I suggest that you check the question again. There might be a typo somewhere.

 

[Poirot1]

Equal (both = 0) if p = q = 1/2
Also equal if a = k, of course.

Sir Poirot, just noticed I received this compliment from you:
"Have I not stated that p+q=1, which you, in your eagerness to be contemptous, have ignored."
May I humbly defend myself by reminding you thay my post says: q=2 and p=-1 : 2 + (-1) = 1.

Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you receive that?

---------- Post added at 15:30 ---------- Previous post was at 15:26 ----------

If it helps. the context is gambler's ruin problem. If you are familiar, then p is the probability of moving up one, q is the probability of moving down one, 'a' is the inital money of player A, and b=k-a is the inital money of B. I am trying to derive the probability of A winning the game, which is apparently the RHS of my equation. The probability B is ruined is the LHS so I thought they would be the same.

 

[Wilmer]

Yes that is why I deleted it immediately (so why are you bringing it up)?. How did you receive that?

I am not at liberty to divulge such, since the incident is presently under investigation.

 

[Poirot1]

By the thought police?

 

[Poirot1]

Wilmer, surely you are allowed to say why you brought it up. You could see I had deleted it. (by the way who on Earth is 'Sir Poirot').