It might be a very silly question, but I just can't figure it out.
Let $X_1,...,X_n$ be random variables with pmfs: $$f(k,p)= \begin{cases} p_1, \hspace{0.5cm} \text{if } k=a\\ p_2, \hspace{0.5cm} \text{if } k=b\\ p_3, \hspace{0.5cm} \text{if } k=c\\ p_4, \hspace{0.5cm} \text{if } k=d \end{cases} $$ where $p_4=1-p_1-p_2-p_3$. I am trying to show that when we have a likelihood function: $$L(\theta|x)=p_1^{y_1}p_2^{y_2}p_3^{y_3}p_4^{y_4}$$ where $y_1,...,y_4$ are counts of $a,b,c,d$ respectively (and $y_4=n-y_1-y_2-y_3$). I want to show that for each $p_i$, the $MLE$ is $\frac{y_i}{n}$, but I can't really do it. So far I have showed that for $MLE$s the following is true: $$\frac{p_1}{y_1}=\frac{p_2}{y_2}=\frac{p_3}{y_3}=\frac{1-p_1-p_2-p_3}{n-y_1-y_2-y_3}$$ Could anyone help me out?
If you let $c$ be equal to your $\displaystyle \frac{\hat p_1}{y_1}=\frac{\hat p_2}{y_2}=\frac{\hat p_3}{y_3}=\frac{1-\hat p_1-\hat p_2-p_3}{n-y_1-y_2-y_3}$ then you will find $$\hat p_1= cy_1$$ $$ \hat p_2=cy_2$$ $$\hat p_3=cy_3$$ $$ 1-\hat p_1-\hat p_2-\hat p_3=c(n-y_1-y_2-y_3)$$ and add these up to give $1=cn$ so $c=\dfrac1n$, which is what you are aiming for