Given two independent random variables $X$ and $Y$ with distribution $X\sim Poisson(\phi)$ and $Y\sim Poisson(2\phi)$, and the observations $X=2$ and $Y=4$ of these.
Show that the expression for log-likelihood function is given by:
$L(\phi)=[4ln(2)-ln(2!)-ln(4!)]+6ln(\phi)-3ln(\phi)$
Plott $L(\phi)$ for values of $\phi \in [0,10]$ (rough scetch). For what values of $\phi$ does $L(\phi)$ have its maximum?
Now, I have written:
Probability mass function, pmf: $P(X=k)=\frac{\lambda^k}{k!}(e^{-\lambda})$
$$EDIT$$
Since $X,Y$ are independent, we have that $X+Y\sim Poisson(\lambda_X +\lambda_Y)$:
\begin{align}\
& P(x=2)*P(y=4)
& = \left(\frac{(\phi)^2}{2!}\right)e^{-\phi}*\left(\frac{(\phi)^4}{4!}\right)e^{-2\phi}
& = \frac{\left(e^{-3\phi}\right)\left(\phi^6\right)}{3} \\
\end{align}
Taking $ln$ of this yields:
\begin{align}\
& ln\left(\frac{\left(e^{-3\phi}\right)\left(\phi^6\right)}{3}\right) \\
& = (-3\phi)ln(e)+6ln(\phi)-ln3 \\
& = (-3\phi)+6ln(\phi)-ln3
\end{align}
I still end up with the wring expression. What am I not getting?
Best Answer
Since $X, Y$ are independent, you can multiply the likelihoods together to get the likelihood for the sample of two observations; the likelihood is $$\frac{e^{-\phi}\phi^2}{2!}\frac{e^{-2\phi}(2\phi)^4}{4!}=\frac{e^{-3\phi}2^4\phi^6}{2!4!}$$
So the logarithm of the likelihood is
$$L(\phi|2,4)=-3\phi+4\log 2+6\log \phi-\log 2!-\log 4!$$
I'm sorry this doesn't exactly match your answer, maybe there was a typo in your solution.