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So, my teacher is teaching about how to choose between estimators in the statistics class, and he proposed the following problem:

We have a random variable X described by a uniform continuous distribuition defined by $$ f_X(x) = \begin{cases} \dfrac 1\theta, & 0 < x < \theta \\ 0, & \text{elsewhere} \end{cases} $$ and we need to find a good estimator for theta that he called $$\theta_3 = max\{X_n\}$$ $$\theta_4 = \frac{n+1}{n}\theta_3 $$ Where Xn is the vector of all the observations x that are i.i.d. He actually have done the calculations for the mean of theta 3: $$E[\theta_3] = \frac{n}{n+1}\theta$$ and from this we have that $$E[\theta_4]=\theta$$ In the end of the class he asked us to calculate the variance of this biased estimator to compare with other estimator that isn't important to my question. My problem is that the formula for calculating the variance is $$ Var(\theta_4) = \int^{\infty}_{-\infty}\ ( x - E[\theta_4] )^2\theta_4 dx = \frac{n+1}{n}\int^{\infty}_{-\infty}\ ( x - E[\theta_4] )^2max\{Xn\} dx $$ and I don't know how to procceed with this max function.

How can I procced with the integral?

StubbornAtom
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Lucka de Godoy Gianvechio
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1 Answers1

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My problem is for calculating the variance of max(x)

As you probabily know, the CDF of $T=max(X_n)$ is the product of the n CDF's

$$F_T(T)=\frac{t^n}{\theta^n}$$

thus its density is

$$f_T(T)=\frac{nt^{n-1}}{\theta^n}$$

Now the problem is very easy remembering that

$$V(T)=E(T^2)-E^2(T)=\frac{n}{\theta^n}\int_0^{\theta}t^{n+1} dt-\Bigg[\frac{n}{n+1}\theta\Bigg]^2=\frac{n}{(n+2)(n+1)^2}\theta$$

Reading what you wrote (there is more than one error), I suggest you to remember also that $V(aX)=a^2V(X)$ thus

$$V(\theta_4)=\frac{(n+1)^2}{n^2}V(\theta_3)$$

tommik
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