I am attempted to procure a function from a composition when given the PDF (I typed the full problem at the bottom in its entirety in case I left out details in my inquiry).
I understand how to get the PDF of a transformed uniformly distributed RV, $X=\phi \circ U$, which can be expression as following: $$\frac{dP_x}{dx} = \frac{du}{dx} = \frac{1}{\phi'(\phi'(x))}$$
However, I am getting confused with regards to reversing this process. The target PDF will be a function of the transformation, I believe. So, after separating the variables and integrating both sides, I would need so solve the following equation (I can't figure out if there are bounds or not, or if I am solving it correctly): $$u=\int_{0}^{+\infty} e^{-x} dx \text{ or }u=\int e^{-x} dx$$
Then, I would need to actually find $x=\phi(u)$ by solving the resulting equation. My next question would be how I would figure out the integration constant in this case?
I believe I am just missing a small concept that would allow me to fully understand how to solve this problem. Any help or guidance in the right direction would be much appreciated!
Full Problem:
Suppose that $U$ is a standard unit random variable that is uniformly distributed on the real interval $[0,1)$. Find a function $\phi(u)$ such that the transformed RV, $X=\phi \circ U$ will have the following probability density function on the interval $[0,+\infty)$: $$p_X(x)=e^{-x}$$
I admit the following response is a bit sloppy, so scrutiny is very much appreciated.
I would say take a guess and suppose we have a one-to-one relationship. Then it must be the case that $$e^{-x} = \frac{f_U(u)}{\left|\frac{dx}{du}\right|} = \frac{1}{\left|\frac{dx}{du}\right|} = \left|\frac{du}{dx}\right|$$ I think you can drop the absolute value since $e^{-x}$ is positive. Hence we have $$du = e^{-x}\,dx\implies\int\,du =\int e^{-x}\,dx\implies u = e^{-x}$$ where I think you can ignore the constant of integration since $e^{-x}$ is a density. Also, it is an indefinite integral since $$\int_0^\infty e^{-x}\,dx = 1.$$ Thus we have that $$X = -\log U.$$
You can check that this is the correct transformation using a one-to-one change of variable, or the cdf approach by calculating $$P(X \leq x) = P(-\log U \leq x).$$