Maximum Likelihood Estimator of a Poisson Distribution

Question

Suppose that independent observations $X_{1}$ and $X_{2}$ are taken from Poisson $P(aλ)$ and Poisson $P(bλ)$ distributions respectively, where $a$ and $b$ are known and positive.

(i) Find the maximum likelihood estimator of $λ$.

My Solution:

$L(λ) = \prod_{i = 1}^{n} P(X_{i} |λ)$

$L(λ) = P(X_{1} |aλ) . P(X_{2} |bλ) $

$L(λ) = \frac{e^{-aλ}λ^{X_{1}}}{X_{1}!}.\frac{e^{-bλ}λ^{X_{2}}}{X_{2}!}$

$L(λ) = \frac{e^{-λ(a+b)}λ^{X_{1} + X_{2}}}{X_{1}!X_{2}!}$

$\log_{e} (L(λ))$ =$ \log_{e} (e^{-λ(a+b)})$ + $\log_{e} (λ^{X_{1} + X_{2}})$ - $\log_{e} (X_{1}!X_{2}!)$

$\frac{d}{dλ} L(λ) = 0$ at maximum

Therefore,

$0 = -(a + b) + \frac{X_{1}X_{2}}{λ}$

$λ = \frac{X_{1} + X_{2}}{a + b}$

Please may someone tell me if this is the correct solution or if I have misunderstood the question?

Answer 1

Your working is fine.

Notice that $\frac{d^2L(\lambda)}{d \lambda^2}=-\frac{X_1X_2}{\lambda}<0$ hence the optimal value is attained at that value.