Is the following likelihood equation for a binomial experiment correct? Question:
Let $N = 100$ and $x = 58$. You can think of this as a binomial experiment with $N$ repetitions and a success probability
$(\theta + 1)/3$. Write down the expression for the likelihood $p(x \mid \theta)$.
My thoughts:
Do I just use the general likelihood equation of:
$$p(x \mid \theta) = {n \choose x}  \theta^x (1 - \theta)^{n-x} $$
and solve the problem?
In this case is it something like this?
$$p(x \mid (\theta + 1)/3) = {100 \choose 58} \left(\frac{\theta + 1}{3}\right)^{58} \left(1 - \frac{\theta + 1}{3}\right)^{100-58} $$
am I doing it correctly? Can anyone please help?
 A: The likelihood function is a function of $\theta$ with $x$ fixed.  In this case it is
$$
L(\theta) = \binom{100}{58}\left(\frac{\theta+1}{3}\right)^{58} \left(1-\frac{\theta+1}{3}\right)^{100-58}.
$$
The term "likelihood equation" usually means either the equation $\dfrac{d}{d\theta}L(\theta)=0$ or the equation $\dfrac{d}{d\theta}\log L(\theta)=0$.  (The set of values of $\theta$ that satisfy the equation is the same either way.)
A: You got confused a bit; the expression $p(x\mid\theta)$ that you wrote down is the probability mass function for the binomial distribution with success probability $\theta$ (not "the general likelihood equation", which isn't really a thing). But your theta is not the success probability, it is instead a parameter, and the success probability depends on it as $\frac{\theta+1}{3}$.
So it makes little sense to write $p(x\mid\frac{\theta+1}{3})$, because you don't condition on the success probability (and if you were, you would be abusing notation too much here), you should condition on the parameter instead; the expression would be $p(x\mid\theta)$ (where $\theta$ is a parameter, not the success probability).
You got the right answer anyway, but you should really pay attention to what you're doing and writing down in the process.
