3
$\begingroup$

Please help me understand the following:

Suppose a tester recorded the quantity $Y=X_1+\cdots X_n$ where $X_i$ has a Poisson distribution with mean $\theta$.

Now, the tester lost all samples $X_i$ and want to create fake observations $Z_i\cdots Z_n$. The tester knows that $$ P(\mathbf Z|Y)=\frac{P(\mathbf Z,Y)}{P(Y)} $$ It is known that $P(Y)$ is also a Poisson distribution with mean $n\theta$ while $P(\mathbf Z,Y)$ is a product of multiple Poisson pdfs with mean $\theta$.


I am asked to find the likelihood function. However, the book hinted that the likelihood is a product of the conditional $P(\mathbf Z|Y)$ and $P(Y)$.

Using that hint, I have shown that the likelihood is the proportional to the likelihood of $X_i\cdots X_n$.

I am confused, why is $l(\theta)=P(\mathbf Z|Y)P(Y)$? This equation is the joint distribution $P(\mathbf Z,Y)$ and I argue that $P(\mathbf Z|Y)$ should be the likelihood by definition.

$\endgroup$

1 Answer 1

4
$\begingroup$

$P(\mathbf Z|Y)$ is the likelihood. I don't know where you found different information, but either it is wrong or you must have misunderstood it. If you search our site, you'll find multiple examples.

$\endgroup$
1
  • $\begingroup$ I had the same thought but $P(\mathbf Z|Y)$ has a Multinomial distribution and I would get a different answer. But I am confident in your answer more than my book, Introduction to Mathematical Statistics (Hogg), which may had typos $\endgroup$
    – wd violet
    Commented May 23, 2023 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.