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I'm trying to solve an exercise from a past exam. The question is:

Suppose there is an urn with 10 balls, where $\theta$ are white and $10 - \theta$ are green. Two balls are then extracted, without replacement. Let $X_i = 1$ denote if the color of the $i$th extracted ball is white, and $X_i = 0$ if green, for $i=1,2$. I) Obtain the maximum likelihood estimator for $\theta$. II) Verify if the estimator obtained is unbiased.

Here is my work so far.

Since there the first and second draw are not independent, the likelihood function is:

$$ \text{L}(\theta) = \text{P}(X_1 = x_1)\cdot\text{P}(X_2=x_2 |X_1 = x_1) $$

For the individual cases of $X_1$ and $X_2$ I can see that:

$$ \text{P}(X_1 = 1) = \frac{\theta}{10} \\ \text{P}(X_1 = 0) = \frac{10 - \theta}{10} \\ \text{P}(X_2= 1|X_1 = 1) = \frac{\theta - 1 }{9} \\ \text{P}(X_2= 0|X_1 = 1) = \frac{ 8 - \theta }{9} \\ \text{P}(X_2= 1|X_1 = 0) = \frac{\theta}{9} \\ \text{P}(X_2= 0|X_1 = 0) = \frac{9 - \theta}{9} \\ $$

I can then develop each case of the likelihood function for the sample, but I can't really go from there. Deriving and equating to zero doesnt give me cases that are present in the parametrer space $\theta \in \{1, \dots, 10\}$. Even with those separate cases, I don't know how evaluate the bias of the estimator when it is not a closed expression.

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    $\begingroup$ $\theta \in \{x_1 + x_2, \dots, 10\}$. It's discrete, and you can just enumerate, or better yet maximize the quadratic likelihood function and check the two points on either side; if it's concave, you know the maximizing integer value is either the floor or ceiling of the maximizing continuous value. In the cases where it isn't concave, the maximizing value will be one of the endpoints, either $0$ or $10$. $\endgroup$
    – jbowman
    Nov 13 '21 at 22:42
  • $\begingroup$ @jbowman This quadratic likelihood function? $\endgroup$ Nov 13 '21 at 23:26
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    $\begingroup$ @Galen - if you work out the likelihood function from the cases given for $X_1$ and $X_2$, it's a quadratic, as it happens. $\endgroup$
    – jbowman
    Nov 14 '21 at 2:37
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The discreteness does make this potentially interesting. Fortunately, there are only four possible outcomes: 2 white, white first, white second, 0 white. It's pretty obvious that if you get 2 white the MLE is $\hat\theta=10$ (because increasing $\theta$ increases the likelihood) and if you get 0 white the MLE is $\hat\theta=0$.

If you get one of each, you could argue by symmetry that if there's a unique MLE it has to be 5/10, but let's actually do the calculation $$L(1,0;\theta)=\frac{\theta\times(10-\theta)}{10\times 9}$$ $$L(0,1; \theta) =\frac{(10-\theta)\times\theta}{10\times 9}$$

So it is the same either way (as it should be). Differentiating, the likelihood is maximised at $\hat\theta=5$, and since that's the maximum over continuous $\theta$ it must also be the maximum over discrete $\theta$.

If you code white as 1 and green as 0, $\hat\theta/10$ is just the sample mean, and we know that's unbiased for the population mean $\theta/10$ under simple random sampling, regardless of the distribution. [Or, it's a straightforward calculation, if you don't want to just know that]


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  • $\begingroup$ I really can't see how the argument by symmetry for the 1,0 and 0,1 cases can apply here, since the two draws are not independent. I think your entire argument will apply, but with different likelihoods and estimates for each sample that can be drawn. $\endgroup$ Nov 15 '21 at 4:35
  • $\begingroup$ I didn't in fact argue by symmetry for t he (0,1) and (1,0) cases; I noted that you could, but actually wrote down the products of conditional probabilities. The reason you can, even though I didn't, is that the entire problem is invariant under swapping the colours on all the balls and changing $\theta$ to $10-\theta$. So if the MLE is $\theta_0$ for (green,white) it must be $10-\theta_0$ for (white,green) $\endgroup$ Nov 15 '21 at 5:16

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