Above is a theorem for how one can find the pdf to the inverse function. I have a problem with the part $P[g(X)\leq y]=P[X \leq h(y)]\: \: \: (a)$
Can someone prove this mathematically? I have tried a little myself I ended up with:
$$P[g(X)\leq Y]=\int_{0}^{g^{-1}(x)}g(x)dx \: \: \: (aa)$$
$$P[h(y)\leq X]=\int_{0}^{h^{-1}(y)}h(y)dy \: \: \: (aaa)$$
Now I am not sure how (aa) and (aaa) are equal. I have also seen some reasoning with the graph of the function one are looking at like the one I have added in the attachment but I dont get this reasoning either. Can someone prove the relation (a) mathematically?
Please tap image to read it.
As a comment to the answer I have addded this:
If $g$ and $h$ are inverses of each other and ar monotonically increasing then:$$g(X)\leq y\implies X=h(g(X))\leq h(y)$$and:
Do you just simply use the invere function on both sides of the equation? How can one prove that using h on both sides of the equation holds?
$$X\leq h(y)\implies g(X)\leq g(h(y))=y$$
So we have $$g(X)\leq y\iff X\leq h(y)$$or equivalently:
How can one just use the equality sign below? I am sure this is related to my first problem. Why would the fraction of outcomes be the same after you have used the inverse?
$$\{g(X)\leq y\}=\{X\leq h(y)\}$$and consequently:$$\mathsf P(g(X)\leq y)=\mathsf P(X\leq h(y))$$
If $g$ and $h$ are inverses of each other and ar monotonically increasing then:$$g(X)\leq y\implies X=h(g(X))\leq h(y)$$and: $$X\leq h(y)\implies g(X)\leq g(h(y))=y$$
So we have $$g(X)\leq y\iff X\leq h(y)$$or equivalently:$$\{g(X)\leq y\}=\{X\leq h(y)\}$$and consequently:$$\mathsf P(g(X)\leq y)=\mathsf P(X\leq h(y))$$