Let $X$ be a scalar random sample of from the following density:

$$f(x|\theta) = \frac{2(\theta-x)}{\theta^2} \quad \quad \quad \text{for } 0 \leq x \leq \theta.$$

Find a pivotal quantity. (I have been given the following hint: Consider the distribution of $Y=\frac{\theta-x}{\theta}$ and use the transformation technique to find the distribution of $Y$.)

Finding the distribution of $Y$, I got $g(y)=2y$. From here, I don't know where to go. I am unsure of how to identify the type of distribution in order to find the information needed for the pivotal quantity, which I believe is found by using $Z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}$ if I am correct?

  • 2
    $\begingroup$ This question calls for the self-study tag. $\endgroup$
    – Xi'an
    Commented Mar 22, 2020 at 7:58

1 Answer 1


It appears that you are confusing yourself by bringing in the pivotal quantity $Z$ that comes from a completely different type of distribution. That quantity does not arise in this problem, since you have only one observation, and the parameters in that pivotal quantity are not defined in this problem.

In general terms, a pivotal quantity is just a function of the observable data and parameters that has a distribution that does not depend on the parameters. So, in this question, once you have shown that $Y$ has a distribution that does not depend on $\theta$, you have shown that $Y$ is a pivotal quantity ---i.e., there is nothing left for you to do. In this case, the "hint" you were given is effectively giving you the pivotal quantity, and all you needed to do was show that its distribution does not depend on $\theta$.

Confirming the pivotal quantity: I am getting the same answer as you for the distribution, but it is a good idea to specify the support of the distribution. For all $0 \leqslant y \leqslant 1$ we have:

$$\begin{equation} \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y \leqslant y) &= \mathbb{P}(\tfrac{\theta-X}{\theta} \leqslant y) \\[6pt] &= \mathbb{P}(X \geqslant (1-y) \theta) \\[6pt] &= \int \limits_{(1-y)\theta}^\theta \frac{2 (\theta-x)}{\theta^2} dx \\[6pt] &= \Bigg[ \frac{x (2 \theta - x)}{\theta^2} \Bigg]_{x=(1-y)\theta}^{x=\theta} \\[6pt] &= 1 - 2 (1-y) + (1-y)^2 \\[6pt] &= 1 -2 + 2y + 1-2y+y^2 \\[6pt] &= y^2. \\[6pt] \end{aligned} \end{equation}$$

This gives the corresponding density function:

$$f_Y(y) = 2y \quad \quad \quad \text{for } 0 \leqslant y \leqslant 1.$$

Since $f_Y$ does not depend on the parameter $\theta$, the function $Y$ is a pivotal quantity in this problem. Note that this quantity has no particular relationship with $Z$, which is the pivotal quantity from an entirely different problem.

Using the pivotal quantity: It might be useful for you to understand that pivotal quantities are used to form confidence intervals. This is done by forming a probability statement on the pivotal quantity and then "inverting" this statement to make it a statement about the location of the parameter of interest. In the present case, for any value $0 < \alpha < 1$ we can form the probability statement:

$$\begin{equation} \begin{aligned} 1-\alpha &= \mathbb{P}(0 \leqslant Y \leqslant \sqrt{1-\alpha}) \\[6pt] &= \mathbb{P}(0 \leqslant 1-\tfrac{X}{\theta} \leqslant \sqrt{1-\alpha}) \\[6pt] &= \mathbb{P}(1-\sqrt{1-\alpha} \leqslant \tfrac{X}{\theta} \leqslant 1 ) \\[6pt] &= \mathbb{P} \Big( X \leqslant \theta \leqslant \frac{X}{1-\sqrt{1-\alpha}} \Big). \\[6pt] \end{aligned} \end{equation}$$

Substituting the observed value $x$ gives the following $1-\alpha$ level confidence interval for $\theta$:

$$\text{CI}_\theta(1-\alpha) = \Big[ x, \frac{x}{1-\sqrt{1-\alpha}} \Big].$$

  • $\begingroup$ Is it a typo: $f(y)$ should be $2y$ instead of $y$ since $F(y)=y^2$? $\endgroup$
    – TrungDung
    Commented Jan 20, 2021 at 11:18
  • $\begingroup$ Do you aware of any trick in general to find a pivotal quantity? $\endgroup$
    – TrungDung
    Commented Jan 20, 2021 at 11:19
  • $\begingroup$ As stated in the answer, $Y$ is the pivotal quantity (its distribution does not depend on $\theta$). $\endgroup$
    – Ben
    Commented Jan 20, 2021 at 12:36
  • $\begingroup$ My question is in general? In general, do we have any strategy to find a pivotal statistic? $\endgroup$
    – TrungDung
    Commented Jan 20, 2021 at 16:40
  • $\begingroup$ That sounds like a useful thing to post as a question on this site. $\endgroup$
    – Ben
    Commented Jan 20, 2021 at 22:21

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