# Why is the likelihood the product of the conditional PDF and the PDF of the parameter

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.

$$P(\mathbf Z|Y)$$ 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.

• 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 Commented May 23, 2023 at 8:38