# Finding density function of random variable

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.

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 $$Y$$.