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.


1 Answer 1


$P(\mathbf Z|Y)$ 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.

  • $\begingroup$ 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

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