Question

Suppose you are playing two unfair coins. The probability of tossing a tail is θ for coin 1, and 2θ for coin 2. You toss each coin for several times, and you get the following results:

(a) What is the probability of tossing a head for coin 1 and for coin 2?

coin no.	result
1	head
2	head
1	tail
1	head
2	tail
2	tail

(b) What is the likelihood of the data given $\theta$?

(c) What is the maximum likelihood estimation for $\theta$?

Answer 1

(a)

Probability of tossing a head for coin 1 = 1 - $\theta$

Probability of tossing a head for coin 2 = 1 - 2$\theta$

(b)

The likelihood of the data given $\theta$ is,

P(head for coin 1) * P(head for coin 2) * P(tail for coin 1) * P(head for coin 1) * P(tail for coin 2) * P(tail for coin 2)

= (1 - $\theta$) * (1 - 2$\theta$) * $\theta$ * (1 - $\theta$) * 2$\theta$ * 2$\theta$

L($\theta$) = 4$\theta^3$ (1 - $\theta$)²(1 - 2$\theta$)

(c)

For maximum likelihood estimation of $\theta$

dL($\theta$)/d$\theta$ = 0   

$\frac{\mathrm{d} L}{\mathrm{d} \theta} = \frac{\mathrm{d} }{\mathrm{d} \theta} 4\theta^3 (1-\theta)^2 (1-2\theta) = 0$

$\Rightarrow 12\theta^2 (1-\theta)^2 (1-2\theta) - 8\theta^3 (1-\theta) (1-2\theta) - 8\theta^3 (1-\theta)^2 = 0$

$\Rightarrow 3(1-\theta) (1-2\theta) - 2\theta (1-2\theta) - 2\theta(1-\theta) = 0$ if $\theta \ne 0$ and $\theta \ne 1$

$\Rightarrow 3 + 6 \theta^2 - 9\theta - 2\theta + 4\theta^2 - 2\theta + 2\theta^2= 0$

$\Rightarrow 12 \theta^2 -13\theta + 3= 0$

$\theta = \frac{13 \pm \sqrt{13^2 - 4*12*3}}{2*12}$

$\theta = 3/8, 1/6$

For $\theta = 3/8$

$L(\theta) = 4 (3/8)^3 (1-3/8)^2 (1-2*3/8) = 0.02059937$

For $\theta = 1/6$

$L(\theta) = 4 (1/6)^3 (1-1/6)^2 (1-2*1/6) = 0.008573388$

Thus, likelihood is maximum for $\theta = 3/8$

maximum likelihood estimation of $\theta$ is 3/8