# Maximum Likelihood Estimator for Multinomial.

Suppose that 50 measuring scales made by a machine are selected at random from the production of the machine and their lengths and widths are measured. It was found that 45 had both measurements within the tolerance limits, 2 had satisfactory length but unsatisfactory width, 2 had satisfactory width but unsatisfactory length, 1 had both length and width unsatisfactory. Each scale may be regarded as a drawing from a multinomial population with density

$$\pi_{11}^{x_{11}} \pi_{12}^{x_{12}} \pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}$$

Obtain the maximum likelihood estimates of the parameters.

I have tried this by the following way:

the likelihood function is

$L=L(\pi_{11},\pi_{12},\pi_{21},(1-\pi_{11}-\pi_{12}-\pi_{21}))$

$=\prod_{i=1}^{50}[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}} \pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}}]$

$=[\pi_{11}^{x_{11}} \pi_{12}^{x_{12}} \pi_{21}^{x_{21}}(1-\pi_{11}-\pi_{12}-\pi_{21})^{x_{22}} ]^{50}$

$=[\pi_{11}^{45} \pi_{12}^{2} \pi_{21}^{2}(1-\pi_{11}-\pi_{12}-\pi_{21})^{1} ]^{50}$

$=\pi_{11}^{2250} \pi_{12}^{100} \pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}$

Taking logarithm of the likelihood function yields,

$L^*=\log L=\log [\pi_{11}^{2250} \pi_{12}^{100} \pi_{21}^{100}(1-\pi_{11}-\pi_{12}-\pi_{21})^{50}]$

$=2250\log [\pi_{11}]+100\log [\pi_{12}]+100\log [\pi_{21}]+50\log (1-\pi_{11}-\pi_{12}-\pi_{21})$

Now taking the first derivative of $L^*$ with respect to $\pi_{11}$

$\frac{\partial L^*}{\partial \pi_{11}}$ $=\frac{2250}{\pi_{11}}-\frac{50}{(1-\pi_{11}-\pi_{12}-\pi_{21})}$

setting $\frac{\partial L^*}{\partial \pi_{11}}$ equal to $0$,

$\frac{\partial L^*}{\partial \hat\pi_{11}}=0$

$\Rightarrow\frac{2250}{\hat\pi_{11}}-\frac{50}{(1-\hat\pi_{11}-\hat\pi_{12}-\hat\pi_{21})}=0$

$\Rightarrow \hat\pi_{11}=\frac{45(1-\hat\pi_{12}-\hat\pi_{21})}{44}$

$\bullet$Are the procedure and estimate of $\pi_{11}$ correct?

$\bullet$I have another question that if it is multinomial then where the term $\binom{n}{x_{11}x_{12}x_{21}x_{22}}=\binom{50}{45,2,2,1}$?

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Consider a positive integer $n$ and a set of positive real numbers $\mathbf p=(p_x)$ such that $\sum\limits_xp_x=1$. The multinomial distribution with parameters $n$ and $\mathbf p$ is the distribution $f_\mathbf p$ on the set of nonnegative integers $\mathbf n=(n_x)$ such that $\sum\limits_xn_x=n$ defined by $$f_\mathbf p(\mathbf n)=n!\cdot\prod_x\frac{p_x^{n_x}}{n_x!}.$$ For some fixed observation $\mathbf n$, the likelihood is $L(\mathbf p)=f_\mathbf p(\mathbf n)$ with the constraint $C(\mathbf p)=1$, where $C(\mathbf p)=\sum\limits_xp_x$. To maximize $L$, one asks that the gradient of $L$ and the gradient of $C$ are colinear, that is, that there exists $\lambda$ such that, for every $x$, $$\frac{\partial}{\partial p_x}L(\mathbf p)=\lambda\frac{\partial}{\partial p_x}C(\mathbf p).$$ In the present case, this reads $$\frac{n_x}{p_x}L(\mathbf p)=\lambda,$$ that is, $p_x$ should be proportional to $n_x$. Since $\sum\limits_xp_x=1$, one gets finally $\hat p_x=\dfrac{n_x}n$ for every $x$.