Area under the PDF of an order statitics

Question

Consider a continuous random variable $$X=\min\{Y_1,Y_2,Y_3\}$$ where $Y_1,Y_2,Y_3$ are iid, non-negative random variables having the same PDF, $f_{Y}(x)$ and CDF $F_{Y}(x)$. The PDF of X is well-known and is given by

$$f_{X}(x)=3(1-F_{Y}(x))^2 f_{Y}(x)$$

I want to verify that $f_{X}(x)$ is a valid PDF, that is $\int_{0}^{\infty} f_{X}(x)=1$. If I want to check the area under $f_{X}(x)$, how should I proceed. That is,

$$\int_{0}^{\infty} f_{X}(x)~dx=3 \int_{0}^{\infty}(1-F_{Y}(x))^2 f_{Y}(x)~dx$$

How should I prove that $$\int_{0}^{\infty}\left(1-F_{Y}(x)\right)^2 f_{Y}(x)~dx=\frac{1}{3}$$ so that $\int_{0}^{\infty} f_{X}(x)=1$. Please throw some bones.

Answer 1

There really is nothing to do. The usual way of showing that the pdf of $X$ is what it is is to first find the cdf, and then differentiate. So before finding the pdf, you knew the pdf.

But if the pdf of $X$ was just given to you, integrate, making the substitution $u=F_Y(x)$. Then $du=f_Y(x)\,dx$, so you are just integrating $3(1-u)^2$.

Answer 2

I going to denote the density by $f_Y(x)$ as its just more natural to me. Then all you have to do is

$$ \int_{0}^{\infty}\left(1-F_{Y}(x)\right)^{2}f_{Y}(x)dx=\int_{0}^{\infty}f_{Y}(x)dx-2\int_{0}^{\infty}f_{Y}(x)F_{Y}(x)dx+\int_{0}^{\infty}f_{Y}(x)F_{Y}^{2}(x)dx $$

Now the first integral is just one. To handle the second and third integrals just use the substitution method, i.e. let $u=F_Y(x)$ so that $du=f_Y(x)dx$

I'm sure you can figure the rest out