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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:

$$ g(u)=f(12-3u)\left|\frac{\mathrm d}{\mathrm du}[12-3u]\right|=\frac{(12-3u)^2}{84}|-3|=\frac{(12-3u)^2}{28},\quad 2<u<5, $$ and $g(u)=0$ otherwise.

Is this correct? If not, where have I gone wrong?

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up vote 1 down vote accepted

Looks good to me. One thing is that if you integrate your $f(x)$ over its support, you will get $27/28$ instead of $1$. Is it $81$ instead of $84$?

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+1 Nice catch about $84$ needing to be $81$. – Dilip Sarwate Nov 16 '12 at 3:31
Thanks, I knew something wasn't right, wrote the problem down wrong. – Cory Gross Nov 16 '12 at 8:36

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