Confusion with maximizing likelihood of binomial parameter $p$

Consider the following data from a sample of a binomial distribution $$X$$ with $$n=2$$ and unknown parameter $$p$$: $$P(X=0)=.175,\ \ P(X=1)=.45,\ \ \text{and} \ \ P(X=2)=.375.$$ My goal is to find the maximum likelihood estimate of $$p.$$

First, I found the likelihood function by multiplying the distributions at each $$X_{i}$$, that is, $$L(p)=\binom{2}{0}p^0(1-p)^2 \cdot \binom{2}{1}p(1-p) \cdot \binom{2}{2}p^2(1-p)^0,$$ which simplifies to $$L(p)=2p^3(1-p)^3.$$ Taking the logarithm of this gives $$l(p)=\log L(p) = 3\log (2p) + 3\log (1-p).$$ Then, differentiating with respect to $$p$$ and setting to $$0$$, we have $$\frac{\partial }{\partial p}l(p)= \frac{3}{p}-\frac{3}{1-p}=0,$$ which finally implies that the estimate of $$p$$ would be $$\hat{p}=\frac{1}{2}.$$ However, this is clearly not correct, since I did not even use the data provided in the calculation. Can anyone provide any help?

Let me start stating that the question is somewhat unusual, because maximum likelihood estimates are usually used to estimate parameters based on data, which in this case should be counts for the three results of the experiment.

With data the usual case your likelihood should be:

$$L(p\mid \text{data}) = \left( (1-p)^2 \right)^{n_0} \left( 2p(1-p) \right)^{n_1} \left( p^2 \right)^{n_2}$$

with $$n_0,n_1,n_2$$ being the counts for $$X=0,X=1,X=2$$.

Then the log-likelihood will be:

$$l(p\mid \text{data}) = 2 n_0 \log(1-p) + n_1\left(\log(2)+ \log(p) + \log(1-p) \right) + 2n_2 \log (p)$$

This is a sum of $$\log(p)$$ and $$\log(1-p)$$ terms which will result in a relatively simple formula for $$p$$ when you solve for $$p$$ after setting the derivative to $$0$$ :

It should be: $$\hat{p} = \frac{2n_2+n_1}{2n_0+2n_1+2n_2}$$

In your case one could search for the maximum of the expected log-likelihood by replacing the counts with their probabilities and solving in the same way.

$$\hat{p} = \frac{2P(X=2)+P(X=1)}{2P(X=0)+2P(X=1)+2P(X=2)} = 0.6$$

, which is the maximum position of $$E(\log(L))$$.

BTW: Your starting equation is the likelihood for observing each outcome exactly once.

• How can you replace the counts (integers) with probabilities ? – StubbornAtom Feb 26 at 7:01
• @StubbornAtom If we observe data $(n_0, n_1, n_2)$ such that $n_x/n = \Pr[X = x]$ for $x \in \{0, 1, 2\}$, where $n = n_0 + n_1 + n_2$, then we immediately find $$\hat p = \Pr[X = 2] + \frac{1}{2} \Pr[X = 1].$$ Thus, the MLE is independent of the sample size, so long as the sample matches the provided probability distribution exactly. The variance of the MLE, however, decreases with the amount of data observed. – heropup Feb 26 at 7:49