Suppose $X$ has PDF $$f_X(x) = \frac{x}{18}, \quad 0 \le x \le 6.$$ Determine the probability distribution of the random variable $Y = 2X+10$ using the distribution function (CDF) method.

Would you kindly tell me if my approach is correct?


$$F_Y(y) = P(Y \leq y) = P(2X+10 \leq y) = P\left(X \leq \frac{y-10} {2}\right)$$

$$ \int^\frac{y-10} {2}_0 \frac{x}{18}\, dx= \frac{(y-10)^2} {144}.$$

I just started learning CDF's. Am I correct on this one?


The result is correct, but realize that for a PDF, for $x \not \in \text{range}, PDF = 0$


$0 \le x \le 6 \implies 0 \le 2x \le 12 \implies 10 \le 2x + 10 \le 22$

$\implies 10 \le y \le 22$

  • $\begingroup$ Thank you. What is the way to find the support for y? $\endgroup$ – wa7d Apr 7 '17 at 4:58
  • $\begingroup$ I've added it. If you find this answer useful, you can upvote and accept it :) $\endgroup$ – Amad27 Apr 7 '17 at 5:29
  • $\begingroup$ Thank you but I'm more interested on the process of finding the support rather than the answer itself $\endgroup$ – wa7d Apr 7 '17 at 17:26
  • $\begingroup$ I added how to find the support $\endgroup$ – Amad27 Apr 7 '17 at 18:58


Your result is correct when $$0 \le \frac{y-10}{2} \le 6.$$ However, what happens if $y$ falls outside this range?

Don't forget to find the pdf by differentiating the CDF back, i.e. $f_Y(y) = F_Y'(y)$.

  • $\begingroup$ can you edit the formatting of your answer? it's a little difficult to read $\endgroup$ – wa7d Apr 7 '17 at 4:48
  • $\begingroup$ @wa7d sorry missed a dollar sign $\endgroup$ – gt6989b Apr 7 '17 at 4:50

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