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Suppose we have a uniform random variable $U$ which is defined on the $[0,1]$ interval. Consider the transformation:$$X=k\times \log(U)$$

How would I go about calculating the interval on which $X$ is defined?

I guess on one hand, the upper limit would be given as $k\times \log(1)=0$ . However, I cannot evaluate $k\times \log(0)$. Any ideas? Thanks!

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The Intervall should $(-\infty,0]$, Since the right limit of log at zero is $-\infty$

Edited due to dsaxton's comment: Note that the probability that U takes any value $c \in [0,1]$ is 0, or in math notation $P(U=c)=0$ for every $c \in [0,1]$

Why do we take limits, (do we?)? First, let me remark that log is bijective strictly increasing function from $(0,1] \to (-\infty,0]$. Using some math one could define the continuous extension of in zero log to be $log(0):= -\infty$ However, in that case, $P(X=-\infty)=P(U=0)=0$, and the value $-\infty$ won't play a role.

However, to determine the image of $X$, we employ the range of log on $(0,1]$ being $(-\infty,0]$. In both intervals, and in the determination of the image of log on $(0,1]$ limits are involved

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  • $\begingroup$ Im not sure I follow..U does take on a value of 0 with probability density =1. This is not the case of the transformation of U $\endgroup$ – ChinG Feb 23 '16 at 16:02
  • $\begingroup$ Thanks for your response, but why do we evaluate the limit of the function at the value instead of the function at the value? $\endgroup$ – ChinG Feb 23 '16 at 16:11
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    $\begingroup$ @ChinG The density of $U$ is defined at one (entirely arbitrarily it turns out) but $P(U = 0) = 0$ so you never have to worry about $\log(0)$. $\endgroup$ – dsaxton Feb 23 '16 at 16:12
  • $\begingroup$ Thanks a lot for the lucid explanation. One final question: why are we involving limits of the function at the value here instead of evaluating the function at the value itself? $\endgroup$ – ChinG Feb 23 '16 at 16:15

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