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# calculus - Approach at maximizing this equation?

$$f(\phi) = \frac{{ n_1\choose x }{ n_2 \choose t-x } \phi^x}{\sum_{u=1}^t { n_1\choose u }{ n_2 \choose t-u } \phi^u}$$

I'm trying to maximize the above function, but I am having trouble.

I start by taking the log of the function and taking the derivative. Doing so gives me

$$\frac{x}{\phi} - \frac{\sum_{u=1}^t { n_1\choose u }{ n_2 \choose t-u } u \phi^{u-1}}{\sum_{u=1}^t { n_1\choose u }{ n_2 \choose t-u } \phi^u}$$

Which after setting equal to 0 gives

$$\frac{\sum_{u=1}^t { n_1\choose u }{ n_2 \choose t-u } u \phi^{u}}{\sum_{u=1}^t { n_1\choose u }{ n_2 \choose t-u } \phi^u} = x$$

At this point I tried playing around with the combinatorial factorials, but it leads me know where.

I know a solution exists (as it's the MLE of a hypergeometric distribution), but I'm unsure how to determine $$\phi$$ given the above ratio.

Looking at the sums, I was initially thinking the trick might be the use of binomial sums, but since $$n_1 \neq n_2$$ I don't think that's possible?

I end up with a similar equation in a related problem:

$$x = \frac{\sum_{u=0}^t u\theta^u /u!(t-u)!}{\sum_{u=0}^t \theta^u /u!(t-u)!}$$

How on earth does one solve these? There must be a simple trick I'm missing...

1. If we have a function $$f(x)=g(x)/h(x)$$, the derivative is $$f'(x)=(g'(x)h(x)-g(x)h'(x))/(h^2(x))$$, which is zero (because we want to maximise/minimise $$f$$) when $$g'(x)h(x)=g(x)h'(x)$$ (assuming that $$h(x)\neq0$$). In the numerator of your sum you have a constant times $$x^{\mathrm{constant}}$$, and in the denominator, you have a complicated looking polynomial in terms of $$x$$. Now just write out $$g'h=gh'$$ as above, since each the numerator and denominator are easily differentiated polynomials, and solve (numerically, if necessary).