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We have a system with $M (M\ge 2)$ random variables. The M variables are related as follows. For each i, 1 to M,

$X_i = I_i+Z$, where $I_i$, Z are independent uniform random variables.

What is the density and cummulative distribution function of $Y_1$, where $Y_1$ is the highest of the $X_2$ to $X_M$, conditional on $X_1=x$?

So in essence, we are after the density and cummulative distribution function of the First Order Statistic, conditional on one of the random variables being observed, in an affiliated system, with the affiliation as defined above.

Below are the steps to where I have reached, could someone please comment on whether I am on the right track and whether any further simplifications are possible?

Lastly, Are there known extensions to other distributions or at least maybe normal and lognormal cases?

Thanks in advance

STEPS TRIED

\begin{eqnarray*} X_{i}=I_{i}+Z \end{eqnarray*} \begin{eqnarray*} f_{X_{i}}\left(x_{i}\right)=\begin{cases} x_{i} & 0\leq x_{i}<1\\ 2-x_{i} & 1\leq x_{i}\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} F_{X_{i}}\left(x_{i}\right)=\begin{cases} \frac{x_{i}^{2}}{2} & 0\leq x_{i}<1\\ 2x_{i}-1-\frac{x_{i}^{2}}{2} & 1\leq x_{i}\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} g\left(y\mid x\right) & = & \left(M-1\right)\left[\frac{F\left(y\right)}{F\left(x\right)}\right]^{M-2}\left(\frac{f\left(y\right)}{F\left(x\right)}\right) \end{eqnarray*} \begin{eqnarray*} \left[\vphantom{\frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}}\because\; f_{Y_{i}}\left(y_{i}\mid Y_{j}=y_{j}\right)\right. & = & \frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}\quad Here,\; Y_{M}\leq Y_{i}\leq Y_{j}\leq Y_{1}\;\; are\; Order\; Statistics.\\ & = & \frac{\left(j-1\right)!}{\left(i-1\right)!\left(j-i-1\right)!}\left\{ \frac{F\left(y_{i}\right)}{F\left(y_{j}\right)}\right\} ^{i-1}\left[\frac{F\left(y_{j}\right)-F\left(y_{i}\right)}{F\left(y_{j}\right)}\right]^{j-i-1}\left.\left(\frac{f\left(y_{i}\right)}{F\left(y_{j}\right)}\right)\right] \end{eqnarray*} \begin{eqnarray*} g\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{y}{x}\right]^{2\left(M-2\right)}\left(\frac{2y}{x^{2}}\right) & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\left(2y-1-\frac{y^{2}}{2}\right)}{\left(2x-1-\frac{x^{2}}{2}\right)}\right]^{M-2}\left\{ \frac{2-y}{\left(2x-1-\frac{x^{2}}{2}\right)}\right\} & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} \left[\vphantom{\frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}}\because\; G_{Y_{i}}\left(y_{i}\mid Y_{j}=y_{j}\right)\right. & \approx & \frac{\int_{-\infty}^{y_{i}}f_{Y_{i},Y_{j}}\left(u,y_{j}\right)du}{f_{Y_{j}}\left(y_{j}\right)}\quad\\ & = & \frac{\left(j-1\right)!}{\left(i-1\right)!\left(j-i-1\right)!}\\ & & \left[\frac{\int_{-\infty}^{y_{i}}\left(F\left(u\right)\right)^{i-1}\left(1-F\left(y_{j}\right)\right)^{M-j}\left(F\left(y_{j}\right)-F\left(u\right)\right)^{j-i-1}f\left(u\right)f\left(y_{j}\right)du}{\left(F\left(y_{j}\right)\right)^{j-1}\left(1-F\left(y_{j}\right)\right)^{M-j}f\left(y_{j}\right)}\right] \end{eqnarray*} \begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{\int_{0}^{y}\left(\frac{u^{2}}{2}\right)^{M-2}uxdu}{\left(\frac{x^{2}}{2}\right)^{M-1}x}\right] & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\int_{1}^{y}\left(2u-1-\frac{u^{2}}{2}\right)^{M-2}\left(2-u\right)\left(2-x\right)du}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}\left(2-x\right)}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*}

\begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{2\int_{0}^{y}u^{2M-3}du}{x^{2M-2}}\right] & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\int_{1}^{y}\left(2u-1-\frac{u^{2}}{2}\right)^{M-2}\left(2-u\right)du}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left[\frac{y^{2M-2}}{x^{2M-2}}\right] & 0\leq y,x<1\\ \left[\frac{\left(2y-1-\frac{y^{2}}{2}\right)^{M-1}}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} \left[\because\; H\left(x\right)=\int_{a}^{x}h\left(t\right)dt\;;\; H'\left(x\right)=h\left(x\right)\right] \end{eqnarray*}

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POTENTIAL ANSWER

PLEASE HELP VERIFY THE STEPS AND IF THERE ARE ALTERNATE WAYS TO SOLVE THIS.

\begin{eqnarray*} X_{i}=I_{i}+Z \end{eqnarray*} \begin{eqnarray*} f_{X_{i}}\left(x_{i}\right)=\begin{cases} x_{i} & 0\leq x_{i}<1\\ 2-x_{i} & 1\leq x_{i}\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} F_{X_{i}}\left(x_{i}\right)=\begin{cases} \frac{x_{i}^{2}}{2} & 0\leq x_{i}<1\\ 2x_{i}-1-\frac{x_{i}^{2}}{2} & 1\leq x_{i}\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} g\left(y\mid x\right) & = & \left(M-1\right)\left[\frac{F\left(y\right)}{F\left(x\right)}\right]^{M-2}\left(\frac{f\left(y\right)}{F\left(x\right)}\right) \end{eqnarray*} \begin{eqnarray*} \left[\vphantom{\frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}}\because\; f_{Y_{i}}\left(y_{i}\mid Y_{j}=y_{j}\right)\right. & = & \frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}\quad Here,\; Y_{M}\leq Y_{i}\leq Y_{j}\leq Y_{1}\;\; are\; Order\; Statistics.\\ & = & \frac{\left(j-1\right)!}{\left(i-1\right)!\left(j-i-1\right)!}\left\{ \frac{F\left(y_{i}\right)}{F\left(y_{j}\right)}\right\} ^{i-1}\left[\frac{F\left(y_{j}\right)-F\left(y_{i}\right)}{F\left(y_{j}\right)}\right]^{j-i-1}\left.\left(\frac{f\left(y_{i}\right)}{F\left(y_{j}\right)}\right)\right] \end{eqnarray*} \begin{eqnarray*} g\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{y}{x}\right]^{2\left(M-2\right)}\left(\frac{2y}{x^{2}}\right) & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\left(2y-1-\frac{y^{2}}{2}\right)}{\left(2x-1-\frac{x^{2}}{2}\right)}\right]^{M-2}\left\{ \frac{2-y}{\left(2x-1-\frac{x^{2}}{2}\right)}\right\} & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} \left[\vphantom{\frac{f_{Y_{i},Y_{j}}\left(y_{i},y_{j}\right)}{f_{Y_{j}}\left(y_{j}\right)}}\because\; G_{Y_{i}}\left(y_{i}\mid Y_{j}=y_{j}\right)\right. & \approx & \frac{\int_{-\infty}^{y_{i}}f_{Y_{i},Y_{j}}\left(u,y_{j}\right)du}{f_{Y_{j}}\left(y_{j}\right)}\quad\\ & = & \frac{\left(j-1\right)!}{\left(i-1\right)!\left(j-i-1\right)!}\\ & & \left[\frac{\int_{-\infty}^{y_{i}}\left(F\left(u\right)\right)^{i-1}\left(1-F\left(y_{j}\right)\right)^{M-j}\left(F\left(y_{j}\right)-F\left(u\right)\right)^{j-i-1}f\left(u\right)f\left(y_{j}\right)du}{\left(F\left(y_{j}\right)\right)^{j-1}\left(1-F\left(y_{j}\right)\right)^{M-j}f\left(y_{j}\right)}\right] \end{eqnarray*} \begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{\int_{0}^{y}\left(\frac{u^{2}}{2}\right)^{M-2}uxdu}{\left(\frac{x^{2}}{2}\right)^{M-1}x}\right] & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\int_{1}^{y}\left(2u-1-\frac{u^{2}}{2}\right)^{M-2}\left(2-u\right)\left(2-x\right)du}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}\left(2-x\right)}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*}

\begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left(M-1\right)\left[\frac{2\int_{0}^{y}u^{2M-3}du}{x^{2M-2}}\right] & 0\leq y,x<1\\ \left(M-1\right)\left[\frac{\int_{1}^{y}\left(2u-1-\frac{u^{2}}{2}\right)^{M-2}\left(2-u\right)du}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} G\left(y\mid x\right)=\begin{cases} \left[\frac{y^{2M-2}}{x^{2M-2}}\right] & 0\leq y,x<1\\ \left[\frac{\left(2y-1-\frac{y^{2}}{2}\right)^{M-1}}{\left(2x-1-\frac{x^{2}}{2}\right)^{M-1}}\right] & 1\leq y,x\leq2 \end{cases} \end{eqnarray*} \begin{eqnarray*} \left[\because\; H\left(x\right)=\int_{a}^{x}h\left(t\right)dt\;;\; H'\left(x\right)=h\left(x\right)\right] \end{eqnarray*}

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