# Transformation of a continuous random variable

If $X \sim U(0,a)$, what's the pdf of $e^{2X}$ ? Is it also $U(0,a)$ ? Or perhaps one has to integrate $e^{2X}$ to get a normalizing constant?

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Why was this downvoted? The language is loose but can be cleaned up. –  Ron Gordon Jan 15 '13 at 22:59
For $t > 0$ , $$P( \exp 2X \leq t ) = P( X \leq \frac{\ln t}{2} )$$ –  ACARCHAU Jan 15 '13 at 23:31
hint: $P(Y\leq y)=P(e^{2X}\leq y)=P(X \leq \frac{\ln y}{2})$ –  jay-sun Jan 15 '13 at 23:34
"Is it also $U(0,a)$?" Have you tried working out what values $e^{2X}$ can take on? What value of $X$ will give you $Y = e^{2X} = 0.1$, say? –  Dilip Sarwate Jan 16 '13 at 2:07
@rlgordonma My downvote (it is not the only one) is because the OP seems not to have put in any effort into even thinking about the question before posting it. –  Dilip Sarwate Jan 16 '13 at 2:09