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.

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 [tex]\theta[/tex] is a function of a random sample for which we have that

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

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

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

[tex] 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[/tex]

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.