I am trying to find the probability density of ln(1/X) which can be rewritten as -ln(X) so to find the probability density I first want to find the distribution function. I am not sure how to do so, can somebody help me with determining this? I am looking at when x is uniformly distributed with 0 < X < 1.
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You have to make the distinction between capital $X$ (the random variable) and (lower case) $x$, a real, which is a value taken by $X$. – Jean Marie Oct 30 '20 at 14:43
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If a continuous random variable $X$ whose support is all $\ge0$ has CDF $F_X$ and PDF $f_X=F_X^\prime$, then $Y=-\ln X$ is a real-valued continuous random variable of CDF$$F_Y(y)=P(Y\le y) =P(X\ge e^{-y})=1-F_X(e^{-y})$$and PDF$$f_Y(y)=F_Y^\prime(y)=e^{-y}f_X(e^{-y}).$$I leave as an exercise working out the support of $Y$ in terms of that of $X$.
callculus42
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J.G.
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$$F_X(x)=x$$
You are looking for the density of
$$Y=-log X$$
Use the definition
$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[-logX \leq y]=\mathbb{P}[X\geq e^{-y}]=1-F_X(e^{-y})=1-e^{-y}$$
In other words
$$Y\sim Exp(1)$$
tommik
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@Mike Pinnsoneault : usually one can write log but obviuosly natural log is meant – tommik Oct 30 '20 at 15:27