# $P[X_1<X_2]$ for independent exponentially distributed $X_1$ and $X_2$

Given $X_1$ and $X_2$ are exponentially distributed with parameter $\lambda_1$ and $\lambda_2$ respectively, we can see $P[X_1 < X_2] = \int_0^{\infty}P[X_2 > x | X_1 = x]P[X_1=x]=\frac{\lambda_1}{\lambda_1+\lambda_2}$.

Given the simplicity of this result is there a way to prove this more intuitively like just from the first principles that $P[X_i \in (x,x+dx)] = \lambda_idx, i=1,2$

• The probability $X_1$ beats $X_2$, (comes in first) given one of them happens in the interval $x$ to $x+dx$, is approximately $\frac{\lambda_1\,dx}{\lambda_1\,dx+\lambda_2 \,dx}$. – André Nicolas Dec 12 '14 at 12:41
• Thanks André for the comment. So, irrespective of the time point we are at, if it were given that the event will happen at that time point, then the conditional probability of the result being $X_1$ is $\frac{\lambda_1}{\lambda_1+\lambda_2}$. Given this has to happen at some time point, this becomes the probability. I get the intuition now. But, can you please help me make this a formal argument. – Discretizer Dec 12 '14 at 13:18
• Could be difficult. It may be that in essence we have to write the usual argument. – André Nicolas Dec 12 '14 at 13:29

My acknowledgements to André for his discussions that helped me figure out this answer. Let $A$ denote the event $min(X_1, X_2) \in (x,x+dx)$.

\begin{eqnarray*} P(X_1 < X_2) &=& \int_0^{\infty}\left[P(X_1 < X_2 | A)P(A)\right] \\ &=& \int_0^{\infty}\left[\frac{\lambda_1dx}{\lambda_1dx+\lambda_2dx}P(A)\right] \\ &=& \frac{\lambda_1}{\lambda_1+\lambda_2}\int_0^{\infty}P(A) \\ &=& \frac{\lambda_1}{\lambda_1+\lambda_2} \end{eqnarray*}

Similar arguments show that, given two coins with probabilities $p_1$ and $p_2$ for head, the probability that the first tail happens simultaneously in a series of infinite tosses is $\frac{(1-p_1)(1-p_2)}{1-p_1p_2}$ without needing to calculate the infinite sums.

the best way to solve this is like this \begin{align}P(X_1<X_2) &=\int^\infty_0P(X_1<X_2|X_1=x)f(x)dx\\ &=\int^\infty_0P(x<X_2|X_1=x)f(x)dx \\ &=\int^\infty_0P(X_2>x|X_1=x)f(x)dx\end{align} because $X_2>x$ and $X_1=x$ are independent, so that

\begin{align}P(X_1<X_2) &=\int^\infty_0P(X_2>x)f(x)dx\\ &=\int^\infty_0(1-1+e^{-\lambda_2x})\lambda_1e^{-\lambda_1x}dx\\ &=\int^\infty_0\lambda_1e^{-\lambda_1x-\lambda_2x}dx\\ &=\frac{\lambda_1}{\lambda_1+\lambda_2}\\ \end{align}

• Note that $P(X_1=x)=0$ for every $x$ hence the four first integrals you wrote are actually zero. (Why add this to a question from 1.5 years ago?) – Did Jun 16 '16 at 6:57
• $P(X_1=x)$is the pdf of X_1, maybe I should write small p? why it is zero? this try to add $X_1=x$ term so that the equation become $P(X_1<X_2, X_1=x)$. the reason to add this is because I found the previous answer is not good enough for me to understand. And when i search google for this question, this one is the first. @Did – Hao Jun 16 '16 at 23:39