Question

The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval (θ, θ + 1), where θ is the true but unknown voltage of the circuit. Suppose that Y1,Y2,...,Yn denotearandomsampleofsuchreadings. a Show that Y is a biased estimator of θ and compute the bias. b Find a function of Y that is an unbiased estimator of θ.

Answer 1

Answer:

Step-by-step explanation:

The complete question is

The reading on a voltage meter connected to a test circuit is uniformly distributed over the interval (θ, θ + 1), where θ is the true but unknown voltage of the circuit. Suppose that Y1,Y2,...,Yn denotearandomsampleofsuchreadings.Let Y be the sample mean. a) Show that Y is a biased estimator of θ and compute the bias. b Find a function of Y that is an unbiased estimator of θ.

Recall that an unbiased estimator Y of a parameter $\theta$ is  a function of a random sample for which we have that

$E[Y] = \theta$. When this is not the case, the quantity $E[Y]-\theta$ is called the biased of the estimator.

Recall that for each i, $Y_i$ is uniformly distributed on the interval $(\theta,\theta+1)$, then $E[Y_i] = \frac{\theta + \theta +1 }{2} = \theta + \frac{1}{2}$.

Then, using the linear property of the expeted value, we have that

$E[Y] = E[\frac{1}{n}\sum_{i=1}^{n} Y_i] = \frac{1}{n}\sum_{i=1}^{n} E[Y_i] = \frac{n (\theta+0.5)}{n} = \theta + 0.5$

So, Y is a biased estimator of [tex]\theta [/tex} and the bias is 0.5.

b) We can easily obtain an unbiased estimator of theta by simply substracting the bias to the biased estimator, that is Y-0.5 is an unbiased estimator of the parameter theta.