Let $\left(x_1,x_2,x_3,x_4,x_5\right)$ be the observed values of a random sample of size $5$ from an exponential distribution with parameter $\beta$. Out of five observed values four are given as $x_1=2$, $x_2=4$, $x_3=5$, $x_4=5$ and if the sample variance is $s^2=1.5$, then a point estimate of $\beta$ is?
In this question a valid method of solving would be to assume a value $x$ for $x_5$ and finding sample variance with one equation and one variable. However, this neglects the first statement about the distribution that is being used and fails to take advantage of the properties that come along with the distribution. What is the significance of that first statement and how does it play a role in the solution?
We assume the random sample to be $\{2,4,5,5,x\}$
Since, we are given the sample variance which is defined to be:
$$s^2 = \frac{1}{n-1}\sum\limits_{i=1}^n{(x-\bar x)^2}$$
$$\bar x=\frac{16+x}{5}$$
Therefore, we get,
$$6 = \left(\frac{6+x}{5}\right)^2 + \left(\frac{4-x}{5}\right) + 2\left(\frac{9-x}{5}\right)^2 + 16\left(\frac{x - 4}{5}\right)^2$$
which implies,
$$150 = 20x^2 -160x +470$$
which implies, $x = 4$ upon solving the quadratic equation.
Now, a suitable point estimate for the parameter $\beta$ here would be the sample mean $\bar x$. Thus, we have,
$$\beta = \bar x = \frac{20}{5} = 4$$