I've been struggling with this math problem and would be grateful for some help:

I have a function Q:

Q = m + r [(1-g) p^k - (1+g) (1-p)^k]

And its integral wrt p:

Int = m p + r/(k+1) [(1-g) p^(k+1) + (1+g) (1-p)^(k+1)]

I have found that as p approaches 0 or 1:

Int/p = Q/(k+1) + C

However, I can't seem to work out the constant C in terms of m, r, g and k.

Can somebody help me out, please?

OK, I managed to work out the constant by pure intuition, trial and error, but I have no idea why - I can't explain it:

As p approaches 0:

Int/p = Q/(k+1) + C, where:

C = k [m - r (1+g)] / (k+1)

I know it's right only because it gives exactly the right result for a variety of values m, r, g and k, but can anyone explain it, or work out a proof for this?

Is it allowed to use an iterative approach?

No, I used numerical methods to show that different combinations of values all fit, but I'm looking for algebraic proof, without substituting actual numbers into the variables, except what is stated, that p tends to zero.

You have a slight problem, in that as written, Int/p diverges as p tends to 0. The terms in Int are mp (and it/p is constant), a multiple of p^{k+1} (which/p is small, assuming k is positive), and a multiple of (1-p)^{k+1}, but this final term is not small for small p, and so it/p will diverge.

However, if we assume that when you took the integral of Q wrt p, you were really doing a definite integral from 0 to p, then you need to subtract the lower limit, and Int should be

(your expression for Int) - (this evaluated at p = 0)

= (your expression for Int) - r(1+g)/(k+1)

And then it's fine. From here it's manipulation, and then a hint of binomial expansion.

We want C = Int/p - Q/(k+1) = [(k+1)Int - pQ]/p(k+1). So...

(k+1)Int - pQ

= (k+1)mp + r[(1-g)p^(k+1) + (1+g)(1-p)^(k+1)] - r(1+g) - mp - rp [(1-g) p^k - (1+g) (1-p)^k]

= kmp + r(1+g)(1-p)^k - r(1+g)

= kmp + r(1+g)[1 - kp + smaller terms] - r(1+g) -- (binomial expansion, where "smaller terms" means order p^2 or higher)

~ kp[m - r(1+g)] -- (ignoring those smaller terms)

So C = this/p(k+1) = the value you claim.

Brilliant, well spotted, and thank you!

Now, by the same approach, what is the constant as p approaches 1?

Assuming that everything else is the same, e.g. that we're still integrating from 0 to p (rather than from p to 1), then the expression

(k+1)Int - pQ = kmp + r(1+g)(1-p)^k - r(1+g)

still holds (because the assumption about p going to 0 wasn't used until later).

And here we can just substitute in p = 1, to get

km - r(1+g)

and hence

C = [km - r(1+g)]/(k+1)

Sorry, in that case we should be integrating from p to 1...

BTW, as you've helped me so much I should tell you that Q is a four-parameter probability distribution, and this result can help to calculate the average tail value above or below a certain threshold value Q. These are the so-called "black swans" of risk - extreme outliers with very big impact but low probability.

Only this time, presumably you would want

Int/(1-p) = Q/(k+1) + C

? If so, then the answer for C is the same, I think. Either by similar manipulation (only a hint less nice, as we're in terms of (1-p) this time), or by waving our hands and using a symmetry argument.


Oh, it was clearly probability, but I was trying to avoid thinking about that because probability never made any sense to me!

Yes, that's right, many thanks.

I love working on risk so probability distributions are a critical part of that, however I've had to devise this "synthetic" probability distribution to capture skewness (g) and kurtosis (k), which we tend to ignore at our peril. Anyway, this has been very helpful so thanks again.

Oops! Not quite the same. I forgot that we had (1-g) on one term and (1+g) on the other. So there is a slight change, and the second C (assuming Int/(1-p) on the LHS) will be

C = k[m + r(1-g)]/(k+1)

A '+' in the middle (because of working at the other end of the interval), and a '-' in the bracket because the p^k term dominates.

Thanks, you might just have avoided the next financial crash.

Or actually since we're now talking about extreme upside outliers, you may have just helped to avoid the next financial boom!

This should be be discussed on the TP forums It should have been kept to pm only. Everyone knows that the official language of TwistyPuzzles is English, and you two are speaking fluent gibberish