My Math Forum Maximum of a binomial

 March 19th, 2017, 08:45 AM #1 Newbie   Joined: Nov 2013 Posts: 29 Thanks: 1 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.
 March 19th, 2017, 09:59 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 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.
 March 19th, 2017, 03:31 PM #3 Newbie   Joined: Nov 2013 Posts: 29 Thanks: 1 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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