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March 19th, 2017, 08:45 AM   #1
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Maximum of a binomial

If X is a binomial random variable, for what value of theta is the probability b(x;n,theta) a maximum?

In their notation theta is the probability of a success and n is the number of trials.

I'm not sure where to start to be honest.
For a given value of the parameters I could say the maximum occurs around mean and do some algebra to find it. But with a parameter it seems somewhere different.
Plus, how would a curve be bigger than another curve? If a curve is greater than another curve for all possible values then it should have a greater sum when integrated over the support but these curves have the same area since they're probabilistic. So it leaves me to think that I need to compare the maximums associated with each given curve. To find the maximums of the maximums. If so, would the answer be 1? Because mean=n*theta and theta is in the interval [0,1].

If only this was a continuous function, I'd be able to differentiate this thing and move on.

I appreciate anyones feed back.
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March 19th, 2017, 09:59 AM   #2
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you can differentiate it,

I'm using $k$ instead of $x$ as is common for discrete values.

$P[k] = \binom{n}{k} \theta^k (1-\theta)^{n-k}$

$\dfrac{\partial}{\partial \theta} P[k] =

\binom{n}{k}\left(k \theta^{k-1}(1-\theta)^{n-k} - \theta^k(n-k)(1-\theta)^{n-k-1}\right) =$

To find the maximum we set this equal to zero as usual to obtain

$k \theta^{k-1}(1-\theta)^{n-k} = \theta^k(n-k)(1-\theta)^{n-k-1}$

$k(1-\theta) = \theta (n-k)$

$k = n\theta$

$\theta = \dfrac {k}{n}$

Now, we note that the maximum value on the right hand side occurs at $k=n$ because $0 \leq k \leq n$

Thus across the entire distribution the maximum of $P[k]$ occurs when

$\theta = 1$, and $P[n]=1$

We can note that the distribution is just mirrored when $\theta = 0$ with $P[0]=1$, this is also a maximum.

So in order to maximize a binomial distribution by adjusting $\theta$ you select $\theta=0$ or $\theta=1$ to obtain a maximum value of 1.
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March 19th, 2017, 03:31 PM   #3
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Thanks! I didn't think it was valid to differentiate a discrete function but in hindsight I can reason why its okay.
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