# Probability $P\{\min{X_1,X_2}\leq X_3\}$?

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?

Thanks.

• 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). – Minus One-Twelfth Mar 5 at 23:17
• Can I just add that your calculations are indeed correct. May I ask, why were you suspicious something might be wrong with your solution? – Stan Tendijck Mar 6 at 2:07
• Because I try To use in other things, and it dose not works. – Monir Mar 7 at 1:20