Probability and Transformation function

Question

Can anyone explain to me how to deal with this question? Many thanks

I tried the following steps:

As $P(Y \le y)=P(\psi(V) \le y)=P(V \le \psi^{-1}(y))=\psi^{-1}(y)=\begin{cases} 1&\text{if}\, y\geq 1\\ 1-p&\text{if}\, 0\le y<1\\ 0&\text{otherwise} \end{cases}$

but then I realised this is meaning less since $\psi^{-1}(y)$ is obviously not a one-to-one function and $\psi^(y)$ may not necessarily be a function. I really don't have a clue about how to move on. Please give me a hand!

Answer 1

In general if $X$ is a random variable with CDF $\mathsf F$ and $V$ has uniform distribution on $(0,1)$ and we prescribe $\Phi_F:(0,1)\to\mathbb R$ by:$$v\mapsto\inf\{y\in\mathbb R\mid\mathsf F(y)\geq v\}$$

then $\Phi_F(V)$ can be shown to have the same distribution as $X$.

This because: $$\Phi_F(v)\leq x\iff v\leq\mathsf F(x)$$so that $$\mathsf P(\Phi_F(V)\leq x)=\mathsf P(V\leq\mathsf F(x))=\mathsf F(x)$$

You can apply this here.

Answer 2

If $\phi$ is defined as $$\phi(V)=\min\!\left(\left\lfloor{\frac{V}{1-p}}\right\rfloor,1\right)$$ then $$ \phi(V)= \begin{cases} 0\;\text{with probability}\;1-p\\[4pt] 1\;\text{with probability}\;p\\[4pt] \end{cases} $$ so $\phi(V)$ has the same distribution as $Y$.