# Find the maximum likelihood estimator of $\theta$

Let $(X_1,Y_1), . . . , (X_n,Y_n)$ be a random sample from the discrete distribution with joint probability mass function $$f_{X,Y} (x,y) = \frac{\theta}4 , \space (x,y) = (0,0)\space or \space(1,1)$$ $$= \frac{2 - \theta}4, \space \space (x,y) = (0,1)\space or \space(1,0),$$

with $0 \le \theta \le 2$.

Find the maximum likelihood estimator of $\theta$.

• Is this homework/self-study? Add the tag and your thoughts on the problem. – Juho Kokkala May 2 '14 at 10:20
• You've posted quite a few routine bookwork questions. Please use the self-study tag for such questions, and read its tag wiki info to see what to include when you ask one. – Glen_b May 5 '14 at 5:28

Let $(X,Y) = (X_{1},Y_{1}), \ldots, (X_{n},Y_{n})$.

The likelihood of your sample is:

$$P((X,Y)|\theta) = \prod_{i=1}^{n} (\delta_{X_{i},Y_{i}}\frac{\theta}{4} + (1-\delta_{X_{i},Y_{i}})\frac{2-\theta}{4})$$

$\delta_{X_{i},Y_{i}} = 1$ if $X_{i} = Y_{i}$ and $= 0$ otherwise.

If $k$ is the number of times $X_{i} = Y_{i}$ in your sample, so:

$$P((X,Y)|\theta) = (\frac{\theta}{4})^{k}(\frac{2-\theta}{4})^{n-k}$$

Maximize that for $\theta$ and you will find that the best estimator for $\theta$ is $\frac{2k}{n}$.

• You handed the home work solution – Aksakal May 2 '14 at 13:55
• Thanks, but a little heads-up would have sufficed, since I was thinking along the same line. – A.Chakraborty May 5 '14 at 20:19
• Great answer. I don't get how you're supposed to answer these type of questions effectively without providing almost the full solution. If you don't want an almost complete solution then be VERY specific with the question. – Glen May 5 '14 at 20:29