Probability: Transform to a arbitrary probability distribution

Question

I need help with this one this is a exercise in when I need to show all the step

Answer 1

You want $$ P(g(X) > y) =\int_y^\infty \lambda\exp (-\lambda u) du = \exp (-\lambda y) $$ but this is also, assuming $g$ is increasing from $[0,1]$ to $\Bbb R^+$: $$ P(X > g^{-1}(y)) = 1-g^{-1}(y) $$that is: $$ g^{-1}(y) = 1- \exp (-\lambda y);\\ g(y) = -\frac{\log(1-y)}\lambda $$and you check that $g$ is increasing from $[0,1]$ to $\Bbb R^+$.

Answer 2

The technique known as "inverse transform sampling" is useful to know.

If $X$ is any continuous random variable with cumulative distribution function $F_X$, $F_X(X)$ has a Uniform distribution on $[0,1]$.

From this, note if you define $F_X^{-1}(x) = \inf\{ c : F_X(x) \geq c \}$ where $c \in (0,1)$, then $F_X^{-1}(U)$ has the distribution of $X$ when $U$ is Uniform on $[0,1]$. (This is just for well defined-ness; if $F_X$ is strictly increasing, you can think of this as just the inverse on $(0,1)$. Drawing a picture makes sense of this inverse noting that cumulative distribution functions are non-decreasing).