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I have been given the following problem:

The probability density function of a random variable X is given by:

  • $f(x;θ) = \dfrac{2(θ−x)}{θ^2}$, if $0< x<θ$, $0$ otherwise*

Find the distribution of $U = X/θ$ and specify its domain where the pdf is non-zero.

In order to solve this do I need to find the estimator $θ$ and divide $2(θ−x)/θ^2$ by the estimator?

From what I have read I can find the estimator by finding the partial derivative of $2(θ−x)/θ^2$ with respect to $θ$ and setting it to $0$?

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  • $\begingroup$ No. You don't need any estimators for this. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 29 '15 at 4:15
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If $0\le u\le 1$ then $$ \Pr(U\le u) = \Pr\left( \frac X \theta \le u\right) = \Pr(X\le \theta u) = \int_0^{\theta u} \frac{2(\theta - x)}{\theta^2} \, dx = 1 - (1-u)^2. $$ Hence $$ f_U(u) = 2(1-u) \text{ if }0\le u\le 1 $$ and $=0$ if $u>1$ or $u<0$.

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  • $\begingroup$ How did you determine the domain of $0≤u≤1$ in the first line of your solution? $\endgroup$ – Lauren Aug 29 '15 at 4:29
  • $\begingroup$ @Lauren Solve $1-1(1-u)^2=0$ and $1-1(1-u)^2=1$ For the second equation you will get two solutions. $u=2$ is a invalid solution, because $F(u)$ has to increase on its domain. This is the left half of the parabola. $\endgroup$ – callculus Aug 29 '15 at 5:13
  • $\begingroup$ @Lauren : You have $0\le x\le \theta$, and that implies $0\le\dfrac x \theta \le 1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 29 '15 at 17:19
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No. It's much simpler than that. If $X$ has PDF $f_X(x;\theta)$, then $U = X/\theta$ is just a scale transformation of $X$, so its PDF is directly given by $$f_U(u;\theta) = \theta f_X(\theta u;\theta).$$ The idea is to calculate this and simplify it accordingly.

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