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Imagine $X_1$,...,$X_n$ are i.i.d. uniformly distributed on the interval [0,1] and $X=\max(a_1X_1,..,a_nX_n)$ and $Y=\max(b_1X_1,..,b_nX_n)$ for some constants $a_1,...,a_n$, $b_1,...,b_n$.(All real, positive numbers)

What is the joint pdf (or cdf) of X and Y?

My idea: We have $f_{X,Y}(x,y)=f_{X\mid Y}(x)f_Y(y)$, Now the latter term can be calculated very easily because we have $F_Y(x)=\prod_{i=1}^n F_{b_iX_i}(x)$. So now we need to calculate the conditional probability;

However, I do not know how to continue here;

Any idea?

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We have $$\Pr(X<x,Y<y){=\Pr(\max(a_1X_1,..,a_nX_n)<x,\max(b_1X_1,..,b_nX_n)<y)\\=\Pr(a_1X_1,..,a_nX_n<x,b_1X_1,..,b_nX_n<y)\\=\Pr(X_1<\min\{{x\over a_1},{y\over b_1}\},\cdots , X_n<\min\{{x\over a_n},{y\over b_n}\})\\=U(\min\{{x\over a_1},{y\over b_1}\})\cdots U(\min\{{x\over a_n},{y\over b_n}\})}$$where $U(x)$ is the cmf of uniform distribution on $[0,1]$. It is not generally easy to find the pmf from the above cmf by differentiation. It is all I got for now!

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  • $\begingroup$ Thank you very much; Ill try to work with it - maybe I can work only with the cdf instead of the pdf; $\endgroup$ – user299124 Nov 29 '18 at 10:03
  • $\begingroup$ But if you have an idea how to get to the pdf let me know $\endgroup$ – user299124 Nov 29 '18 at 10:04
  • $\begingroup$ You're welcome. If you felt more details i'll try to make some calculations but the results are easy to be derived only for some particular cases :) $\endgroup$ – Mostafa Ayaz Nov 29 '18 at 10:04
  • $\begingroup$ Actually you must divide each $\min\{{x\over a_i},{y\over b_i}\}$ to the sections where $0<{x\over a_i}<1$ or $0<{y\over b_i}<1$. Also notice that the cdf is nonzero only if $0<\min\{{x\over a_i},{y\over b_i}\}<1$ for all $i$ $\endgroup$ – Mostafa Ayaz Nov 29 '18 at 10:06

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