# Can you find the maximum likelihood, from a probability not conditioned on a random variable?

I have the following distribution, $$p(Y_i=y_i\mid \nu_i, 1000) = \hbox{Poisson} (y_i \mid\lambda \cdot \frac{\nu_i}{1000}) \quad \hbox{ for } \quad i=1,...,n$$

I am wondering if I can find the likelihood function of $$\lambda$$ in this instance. In the case where we have $$p(Y_i=y_i\mid\lambda) = \hbox{Poisson} (y_i \mid\lambda) \quad \hbox{ for } \quad i=1,...,n.$$ it is apparent that the likelihood takes the form of $$L(\lambda)=L(\lambda |y_1,y_2,..,y_n)=P(Y_1=y_1|\lambda)P(Y_2=y_2|\lambda)...P(Y_n=y_n|\lambda)\\[1.1em] =\frac {e^{-\lambda}\lambda^{y_1}}{y_1!}\cdot\frac {e^{-\lambda}\lambda^{y_2}}{y_2!}\cdot...\cdot\frac {e^{-\lambda}\lambda^{y_n}}{y_n!}=\frac{e^{-n\lambda}\cdot\lambda^{\sum_{i=1}^{n}y_i}}{\prod_{i=1}^{n}(y_i!)}$$

but in the second situation it seems that it would take a completely new form, would the first situation take the likelihood form of

$$L(\lambda)=p(Y_1=y_1\mid \nu_1, 1000)p(Y_2=y_2\mid \nu_2, 1000)...p(Y_n=y_n\mid \nu_n, 1000)\\[1.1em] = \frac {e^{-\lambda*\frac{\nu_n}{1000}}(\lambda*\frac{\nu_1}{1000})^{y_1}}{y_1!}\cdot\frac {e^{-\lambda*\frac{\nu_2}{1000}}(\lambda*\frac{\nu_2}{1000})^{y_2}}{y_2!}\cdot...\cdot\frac {e^{-\lambda*\frac{\nu_n}{1000}}(\lambda*\frac{\nu_n}{1000})^{y_n}}{y_n!}$$

this would seem to take the following form:

$$L(\lambda)=\frac{e^{\frac{-\lambda}{1000}\cdot\sum_{i=1}^n \nu_i}(\frac{\lambda}{1000})^{\sum_{i=1}^n y_i}\prod_{i=1}^n v_i^{y_i}}{\prod_{i=1}^{n} y_i!}$$

and then to find the max likelihood of course we would take the ln and then derivative and solve for lambda when it is equal to zero, however is my assumption of the likelihood correct for the second distribution?

• Could you add some more explanation what exactly do you mean? I'm not sure if I understand what you mean by this notation and what exactly is your question?
– Tim
Oct 17, 2020 at 22:58
• Try rearranging terms in your second situation, and you will see (I hope) that the two are not so dissimilar after all. (For an example of what I mean by "rearranging terms", note that $e^{-\lambda v_1/1000}\cdot e^{-\lambda v_2/1000}\cdots e^{-\lambda v_n/1000} = e^{-\lambda \cdot (\sum_i v_i)/1000}$.) Oct 17, 2020 at 23:07
• I updated it @jbowman did I make any errors? Oct 17, 2020 at 23:17
• Now you can get rid of all the terms that don't involve $\lambda$ (in both variants), e.g., the denominator, the $/1000$ term in $({\lambda \over 1000})^{\Sigma \dots}$, etc. Then take the logs and compare the two expressions. Oct 19, 2020 at 17:50