Let $X_1, X_2$ and $X_3$ exponential random variables with the same parameter $\beta$?

The PDF and CDF of $X_i$ are $$ f_{X_i}(x)=\beta e^{-x \beta}, $$ $$ F_{X_i}(x)=1-e^{-x \beta}. $$

The PDF and CDF of $Y=\min\{X_1,X_2\}$ are $$ f_{Y}(y)=2\beta e^{-2\beta y}, $$ $$ F_{Y}(y)=1-e^{-2\beta y}. $$ I would lieke to find the probability that $$ P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}, $$ and the probability $$ P\Big\{\min\{X_1,X_2\}> X_3\Big\}. $$ I use the following steps \begin{align} P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}&=P\Big\{Y\leq X_3\Big\}\\ &=\int_{x_3=0}^{\infty}\left(\int_{y=0}^{x_3}f_Y(y)dy\right)f_{X_3}(x_3)dx_3\\ &=\int_{x_3=0}^{\infty}F_y(x_3)f_{X_3}(x_3)dx_3\\ &=1-\int_{x_3=0}^{\infty}e^{-2\beta x_3}\beta e^{-\beta x_3}dx_3\\ &=1-\beta\int_{x_3=0}^{\infty}e^{-3\beta x_3}dx_3\\ &=1-\frac{\beta}{3\beta}\\ &=2/3. \end{align} For $$ P\Big\{\min\{X_1,X_2\}> X_3\Big\}=1-P\Big\{\min\{X_1,X_2\}\leq X_3\Big\}=1/3. $$

IS my derivation true?


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    $\begingroup$ Incidentally, there is in fact an easier way to obtain the answer: you effectively want the probability that $X_3$ is either "first" or "second" if we sort in order. By symmetry, $X_3$ is equally likely to be "first", "second", or "third", so there is a 2 in 3 chance that it is first or second. In other words, the answer is $2/3$. This will be true as long as are $X_1,X_2, X_3$ are iid (they don't have to be exponentially distributed). $\endgroup$ – Minus One-Twelfth Mar 5 at 23:17
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    $\begingroup$ Can I just add that your calculations are indeed correct. May I ask, why were you suspicious something might be wrong with your solution? $\endgroup$ – Stan Tendijck Mar 6 at 2:07
  • $\begingroup$ Because I try To use in other things, and it dose not works. $\endgroup$ – Monir Mar 7 at 1:20

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