Let $n \in \mathbb Z$. An urn contains three balls, one has the number $n$ and the two others have the number $n+3$. One ball is drawn randomly and it has the number X. On the basis of X, n is to be estimated.

Now we have $\Theta := \mathbb Z, \mathcal X:=\{n,n+3\}, \mathbb P(X=k)=\frac{1}{3}\mathbf 1_{\{k=n\}}+\frac{2}{3}\mathbf 1_{\{k=n+3\}}$.

I don't know how to calculate the maximum likelihood estimator for n here. More precisely, I don't know how to formulate the likelihood function in this case. Any help is greatly appreciated!


1 Answer 1


The likelihood function is equivalent to the probability of seeing $X = x$ given that $n = \eta$, and is given by

\begin{align*} L(\eta \, | \, x) & = \mathbf{P}[ X = x \, | \,n = \eta] \\ & = \frac{1}{3}\mathbf{1}(x = \eta) + \frac{2}{3} \mathbf{1}(x = \eta + 3). \end{align*}

The maximum likelihood estimate $n^*$ is then the choice of $\eta$ which maximizes the above function. Note in particular that the above equation means

$$L(\eta \, | \, x) = \begin{cases} \frac13 & \text{if $\eta = x$,} \\ \frac23 & \text{if $\eta = x-3$,} \\ 0 & \text{else.} \end{cases} $$

Hence the MLE is given by $n^* = x - 3$.


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