We have two independent random variables $X$ and $Y$ with $X\sim Poisson(\Phi)$ and $Y\sim Poisson(2\Phi)$, and the observations $x=2$ and $y=4$ of these.
Show that the expression for the log-likelihood function is given by:
$l(\phi)=[4ln(2)-ln(2!)-4ln(4!)]+6ln(\phi)-3ln(\phi)$
I know that the function for Poisson distribution is given by:
Probability mass function: $P(X = k) = \frac{\lambda ^k}{k!}e^{-\lambda}$
Standard Normal Distribution: $Z=\frac{X-\lambda}{\sqrt{\lambda}} $
Log Likelihhod: $ln[f(x_1,x_2,...,x_n;p)]$
But how do I use these to show the proof?
pmf of a poisson $Po(\phi)$ is
$$P(X=x)=\frac{e^{-\phi}\phi^x}{x!}$$
but as the likelihood depends on the parameter $\phi$ we can say also that
$$L(\theta)\propto e^{-\phi}\phi^x$$
Thus your likelihood becomes
$$L(\phi)\propto e^{-\phi}\phi^2e^{-2\phi}(2\phi)^4$$
taking its log, after some easy algebraic manipulations you get
$$l(\phi)=-3\phi+6\log\phi$$
This expression is equivalent to the one you are requesting to show as loglikelihoods are equivalent but an additive constant (the expression in your [ ] brackets)
Of course there is an evident typo in your statement: your $-3\log(\phi)$ is evidently $-3\phi$... mine is correct!