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Choose an uniformly distributed random variable $U$ on the unit interval $[0,1]$. Then, what is the probability density function of $Y= \ln(U+ 1)$?

I know the density function is the derivative of the distribution function, but im not sure how to solve it like this.

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If $X$ has PDF $f$, then for any measurable function $g$,

$$E(g(X))=\int_\Bbb R g(x)f(x)\mathrm dx$$

Conversely, if you can write $E(g(X))$ in such a form, then $f$ is the PDF of $X$.


Let $g$ be an arbitrary measurable function.

$$E(g(Y))=E(g(\log(1+U)))=\int_{\Bbb R} g(\log(1+t))\mathrm f_U(t) dt=\int_0^1 g(\log(1+t))\mathrm dt$$

Using the change of variable $u=\log(1+t)$,

$$E(g(Y))=\int_0^{\log2} g(u)e^u\mathrm du$$

Hence the PDF of $Y$ is:

$f_Y(t)=e^t\mathbb1_{[0,\log2]}(t)$$


Another approach, with the CDF of $U$, which is $P(U\leq x)=x$ for $x\in [0,1]$.

Then since $t\to\log(1+t)$ is continuous and increasing on $[0,1]$ (hence bijective), for $y\in[0,\log2]$, $P(Y\leq y)=P(\log(1+U)\leq y)=P(U\leq e^y-1)=e^y-1$.

Just take the derivative of this to get the PDF of $Y$.

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