There are 3 types of flowers that can grow from planting a seed. $$P(\text{Daisy}) = \theta_1$$ $$P(\text{Rose}) = (1-\theta_1)\theta_2$$ $$P(\text{Sunflower}) = (1-\theta_1)(1-\theta_2)$$
the total number of flowers at the end is $n.$ If $X=(X_1, X_2, X_3)$ is the number of daisies, roses and sunflowers respectively, find the MLEs for $\theta_1$ & $\theta_2$ and $E[X_1]$ and $E[X_2]$ in term of the parameters $\theta_1$ & $\theta_2$.
I am aware that this is a multinomial distribution but i sont know how to go about finding the estimators for the parameters since there are 2 of them.
The likelihood for the outcome $(X_1,X_2,X_3)=(x_1,x_2,x_3)$ is
$$ \binom n{x_1,x_2,x_3}\theta_1^{x_1}(1-\theta_1)^{x_2}\theta_2^{x_2}(1-\theta_1)^{x_3}(1-\theta_2)^{x_3}\;, $$
so the derivative of the log likelihood with respect to $\theta_2$ is
$$ \frac{x_2}{\theta_2}-\frac{x_3}{1-\theta_2}\;. $$
Setting this to $0$ yields the MLE for $\theta_2$, $\frac{x_2}{x_2+x_3}$, which shouldn’t surprise us, since we effectively have a subexperiment with $x_2$ of the $x_2+x_3$ big flowers being roses, where we don’t care about the daisies.
The same calculation for $\theta_1$ yields
$$ \frac{x_1}{\theta_1}-\frac{x_2+x_3}{1-\theta_1} $$
and thus the MLE $\frac{x_1}{x_1+x_2+x_3}=\frac{x_1}n$ for $\theta_1$, again not surprising, since for the proportion of daisies we don’t care about the distinction between the two types of big flowers.