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$X,Y$ are independent, identical distributed with $$P(X=k) = P(Y=k)=\frac{1}{2^k} \,\,\,\,\,\,\,\,\,\,\,\, (k=1,2,...,n,...)$$

Calculate the probabilities $P(\min\left\{X,Y\right\} \leq x)$ and $P(\max\left\{X,Y\right\} \leq x)$

For the minimum I do like this:

$$\begin{split}F_M(x) &= P(\min\left\{X,Y\right\} \leq x) \\ &= 1-P(x<\min\left\{X,Y\right\} ) \\ &= 1-P(x<X, x<Y) \\ & = 1-P(X>x)\,P(Y>x)\\ & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\ & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}$$

Is this correct for minimum? I'm not sure how do it for $\max$? Maybe I do it too complicated because they are equal these $P(X=k)=P(Y=k)$ maybe you can do it more elegant? But I don't know how?

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  • $\begingroup$ @Harry49 Thank you for saying I do for $\min$ correct! :) Maybe you can make answer pls for the $\max$? I'm a bit confused because of other answer? $\endgroup$
    – eyesima
    Dec 17 '17 at 12:11
  • $\begingroup$ @conime Harry did give you the answer for $\max$. $\endgroup$ Dec 17 '17 at 12:25
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$$\begin{split}F_{\min}(x) &= P(\min\left\{X,Y\right\} \leq x) \\[0.5ex] &= 1-P(x<\min\left\{X,Y\right\} ) \\[0.5ex] &= 1-P(x<X, x<Y) \\[0.5ex] & = 1-P(X>x)\,P(Y>x)\\[0.5ex] & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\[0.5ex] & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}$$

Is this correct for minimum? I'm not sure how do it for $\max$?

It is correct, and for max it is even simpler.

$$\begin{split}F_{\max}(x) &= P(\max\left\{X,Y\right\} \leq x) \\[0.5ex] &= P(X\leq x, Y\leq x) \\[0.5ex] & = P(X\leq x)\,P(Y\leq x)\\[0.5ex] & = F_X(x)\,F_Y(x)\end{split}$$

Also, recall the expansion of a Geometric Series.

$$\displaystyle F_X(x)~{=F_Y(x)\\[0.5ex]= (\sum_{k=1}^x {2}^{-k})\;\mathbf 1_{x\in\{1,2,\ldots\}} \\[0.75ex]= (1-{2}^{-x})\;\mathbf 1_{x\in\{1,2,\ldots\}}}$$

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Yes. It is. You can do the same for the maximum:

$$\mathbb{P}(\max(X,Y)\leq x) = \mathbb{P}(X\leq x) \times \mathbb{P}(Y \leq x) $$

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  • $\begingroup$ pls don't give him down vote he try for help also!!!! $\endgroup$
    – eyesima
    Dec 17 '17 at 12:59

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