I'm confused by a question I need to solve. I'm given the probability function
$$P(Y=y,\theta)=f(y,\theta)=\frac{(\theta t)^y}{y!}e^{-\theta t}, y = 0,1,2,\dots;\theta>0$$
Is it true that $P$ is also the likelihood function?
If also given 20 separate intervals for $Y$, is it true that the maximum likelihood is given by $\frac{\partial P}{\partial\theta}=0$? Since the solution is $\theta_{max}=\frac{y}t$, where $t$ is time, and each interval is given as one hour, then is it correct that $\theta_{max}=y_{\max}$ from the data set?
Just wanted to make sure I'm getting this right. Would appreciate some feedback.
If I'm interpreting you correctly you are given $20$ independent samples that are Poisson-distributed with mean $\theta t$ (where $t$ is the same for each one - representing the duration of some interval of time disjoint from all the others during which you measure the number of clicks of a Poisson process) and you're trying to get the MLE for $\theta.$ The joint distribution for $n=20$ samples is just the product cause they're independent $$ P(Y_1=y_1,\ldots, Y_n=y_n;\theta) = \frac{1}{\prod_i y_i!}(\theta t)^{\sum_i y_i}e^{-n\theta t}.$$ This thing, when viewed as a function of $\theta,$ is the likelihood function.
To get the MLE you maximize this. When you do so by taking the derivative and setting equal to zero, you will get $\theta_{MLE} = \frac{1}{nt} \sum_i y_i.$
I can't follow how you got $\theta_{MLE} = y_{max}$ (also, what is $y_{max}$? The largest $y$?) but it's incorrect.