The probability function of a random variable $X$ is given by
$ f(x) = \left\{ \begin{array}{ll} x^2/81, & \quad -3 < x <6, \\ 0, & \quad \text{otherwise}. \end{array} \right. $
Find the probability density for $ U = \frac{1}{3}(12-X) $.
OK, so I have attempted this problem fully and have come up with an answer. However, I cannot find a problem similar to this one worked out anywhere so I am having trouble convincing myself that I am 100% correct. Here is my solution:
Because $ U = \frac{1}{3}(12-X) $, the inverse transformation will then be $ X = 12-3U $ and when $ -3 < x < 6 $ we would have $2 < u < 5$.
Using the the change of variable technique we plug into the formula to obtain:
$ g(u)=f(12-3u)\left|\frac{\mathrm d}{\mathrm du}[12-3u]\right|=\frac{(12-3u)^2}{84}|-3|=\frac{(12-3u)^2}{28},\quad 2 and $g(u)=0$ otherwise.
Is this correct? If not, where have I gone wrong?