Given two random variables $x,y$ uniformously distribuited on $[0,1]$, define their linear combination $k = ax + by$ with $a>b$.
By convolution, I have derived the CDF and PDF of k as follow:
$$
F(k)= \begin{cases}
\frac{k^2}{2ab} & 0\leq k\leq b \\
\frac{2k-b}{2a} & b < k\leq a \\
1-\frac{(a+b-k)^2}{2ab} & a<k\leq a+b
\end{cases}
$$
$$
f(k)= \begin{cases}
\frac{k}{ab} & 0\leq k\leq b \\
\frac{1}{a} & b < k\leq a \\
\frac{(a+b-k)}{ab} & a<k\leq a+b
\end{cases}
$$
I'm quite confindent on that PDF because by integrating it across $[0,(a+b)]$ I get 1 as a result.
Then I computed the PDF of the maximum order statistic of $k$ (2 draws). The formula I used is:
$$
f_1^{(2)}(k)=2f(k)F(k)
$$
Hence I obtained:
$$
f_1^{(2)}(k)= \begin{cases}
\frac{k^3}{a^2b^2} & 0\leq k\leq b \\
\frac{2k-b}{a^2} & b < k\leq a \\
\frac{(a+b-k)(a^2+b^2+k^2-2ka-2kb)}{a^2b^2} & a<k\leq a+b
\end{cases}
$$
QUESTION. I want to find the PDF of x and the expected value of x within the maximum order statistic of k.
$(x,y)$ are the characteristics of a person. And $k$ is a summatory statistics which allow to rank people. Hence the PDF of the maximum order statistic of k is the probability that someone is ranked first. I need to know the probability that a person with characteristic $"x"$ in the first r.v. is ranked first.
I thought that it was enough to replace $k = ax + by$ within $f_1^{(2)}(k)$ and then look for the marginal distribution of $x$ within $f_1^{(2)}(k(x,y))$. However it does not work. And, I think, the reason is the following (and I find it weird).
If I integrate $f_1^{(2)}(k)$ across all the domain of $k$ I obtain, as expected, 1 as a result:
$$
\int_{0}^{b} \frac{k^3}{a^2b^2}\, dk + \int_{b}^{a} \frac{2k-b}{a^2}\, dk + \int_{a}^{(a+b)} \frac{(a+b-k)(a^2+b^2+k^2-2ka-2kb)}{a^2b^2}\, dk = 1
$$
However If I replace $k = ax + by$ within $f_1^{(2)}(k)$ and I double-integrate within the domain of $x$ and $y$, I do not obtain 1. Why?
$$
\int_{0}^{1} \int_{0}^{\frac{b}{a}(1-y)} \frac{(ax+by)^3}{a^2b^2}\, dx dy+
\int_{0}^{1} \int_{\frac{b}{a}(1-y)}^{1-\frac{b}{a}} \frac{2(ax+by)-b}{a^2}\,dx dy + \int_{0}^{1} \int_{1-\frac{b}{a}}^{1} \frac{(a+b-(ax+by))(a^2+b^2+(ax+by)^2-2(ax+by)a-2(ax+by)b)}{a^2b^2}\,dx dy = \frac{1}{a}+\frac{b}{3a^2}
$$
THANKS!!!
I think I solved the problem, but I hope someone can check my solution. As BGM pointed out, $f_k(ax+by)$ is NOT the joint PDF of $(x,y)$, which are i.i.d. .
What I'm looking is simply the expected value of $x$ conditional to $(ax+by) = E[k_1^{(2)}]$, which can be denoted with $E[x|(ax+by) = E[k_1^{(2)}]]$. Let define this value, with a slight abuse of notation, $E[x_1]$.
Since $(x,y)$ are i.i.d., the PDF of $x_1$ remains uniform; however $x_1$ is no longer (necessarily) distribuited across $0,1$ i,e, the support changes. Suppose in fact that $b<E[k_1^{(2)}]$; then, the lower bound of the support of $x_1$ would be $\tfrac{E[k_1^{(2)}]-b}{a}$. Suppose instead that $a>E[k_1^{(2)}]$; then the upper bound of the support of $x_1$ is no longer $1$ but $\tfrac{E[k_1^{(2)}]}{a}$.
One can work out all the cases, depending on the parameters of the problem.