How to prove $P\{min({X}_{1},{X}_{2},\dots ,{X}_{n})={X}_{i}\}=\frac{{\lambda}_{i}}{{\lambda}_{1}+\cdots +{\lambda}_{n}}$ , when ${X}_{i}$ is exponentially distributed

steveo963200054
2022-09-14
Answered

How to prove $P\{min({X}_{1},{X}_{2},\dots ,{X}_{n})={X}_{i}\}=\frac{{\lambda}_{i}}{{\lambda}_{1}+\cdots +{\lambda}_{n}}$ , when ${X}_{i}$ is exponentially distributed

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Francis Blanchard

Answered 2022-09-15
Author has **12** answers

Assuming the $\text{}{X}_{j}\text{}$ are independent,

$$\begin{array}{rcl}P(\phantom{\rule{thinmathspace}{0ex}}min({X}_{1},{X}_{2},\dots ,{X}_{n}\phantom{\rule{thinmathspace}{0ex}})={X}_{i})& =& P({X}_{i}\le {X}_{j}\text{}{\textstyle \text{for}}j\ne i)\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}P(t\le {X}_{j}\text{}{\textstyle \text{for}}j\ne i|{X}_{i}=t){\lambda}_{i}{e}^{-{\lambda}_{i}t}dt\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}\prod _{j\ne i}P(t\le {X}_{j}){\lambda}_{i}{e}^{-{\lambda}_{i}t}dt\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}{\lambda}_{i}{e}^{-\sum _{j=1}^{n}{\lambda}_{j}t}dt\\ & =& \frac{{\lambda}_{i}}{\sum _{j=1}^{n}{\lambda}_{j}}\text{}.\end{array}$$

$$\begin{array}{rcl}P(\phantom{\rule{thinmathspace}{0ex}}min({X}_{1},{X}_{2},\dots ,{X}_{n}\phantom{\rule{thinmathspace}{0ex}})={X}_{i})& =& P({X}_{i}\le {X}_{j}\text{}{\textstyle \text{for}}j\ne i)\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}P(t\le {X}_{j}\text{}{\textstyle \text{for}}j\ne i|{X}_{i}=t){\lambda}_{i}{e}^{-{\lambda}_{i}t}dt\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}\prod _{j\ne i}P(t\le {X}_{j}){\lambda}_{i}{e}^{-{\lambda}_{i}t}dt\\ & =& \underset{0}{\overset{\mathrm{\infty}}{\int}}{\lambda}_{i}{e}^{-\sum _{j=1}^{n}{\lambda}_{j}t}dt\\ & =& \frac{{\lambda}_{i}}{\sum _{j=1}^{n}{\lambda}_{j}}\text{}.\end{array}$$

cuuhorre76

Answered 2022-09-16
Author has **1** answers

You forgot the assumption ${X}_{1},{X}_{2},\dots ,{X}_{n}$ are independent.

So for t>0 and $\delta t$ small,

$(0,\mathrm{\infty})$

Thus summing partition of $(0,\mathrm{\infty})$ into intervals of length $\delta t$, and taking $\delta t\to 0$,

$$\mathbb{P}(min({X}_{1},\dots ,{X}_{n})={X}_{i})={\int}_{0}^{\mathrm{\infty}}{\lambda}_{i}{e}^{-({\lambda}_{1}+\cdots +{\lambda}_{n})t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=\frac{{\lambda}_{i}}{{\lambda}_{1}+\cdots +{\lambda}_{n}}$$

So for t>0 and $\delta t$ small,

$(0,\mathrm{\infty})$

Thus summing partition of $(0,\mathrm{\infty})$ into intervals of length $\delta t$, and taking $\delta t\to 0$,

$$\mathbb{P}(min({X}_{1},\dots ,{X}_{n})={X}_{i})={\int}_{0}^{\mathrm{\infty}}{\lambda}_{i}{e}^{-({\lambda}_{1}+\cdots +{\lambda}_{n})t}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=\frac{{\lambda}_{i}}{{\lambda}_{1}+\cdots +{\lambda}_{n}}$$

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solves the differential equation

However, this is an error and this y does not solve the differential equation. Is there a simple typo that makes the problem workable?