To solve this, we need to calculate $P(X_1 < X_2)$ where $X_1$ and $X_2$ are independent and exponentially distributed random variables with parameters $\theta_1$ and $\theta_2$, respectively.

For exponential random variables:

• The probability density function (pdf) of $X_1$ is $f_{X_1}(x) = \theta_1 e^{-\theta_1 x}$ for $x \geq 0$.
• The probability density function of $X_2$ is $f_{X_2}(x) = \theta_2 e^{-\theta_2 x}$ for $x \geq 0$.

Step 1: Express $P(X_1 < X_2)$

We want to find: $P(X_1 < X_2) = \int_0^{\infty} P(X_1 < X_2 | X_2 = y) f_{X_2}(y) \, dy.$

Step 2: Calculate $P(X_1 < X_2 | X_2 = y)$

Given $X_2 = y$, we want $P(X_1 < y)$: $P(X_1 < y) = \int_0^y f_{X_1}(x) \, dx = \int_0^y \theta_1 e^{-\theta_1 x} \, dx.$ Evaluating this integral: $P(X_1 < y) = \left[-e^{-\theta_1 x}\right]_0^y = 1 - e^{-\theta_1 y}.$

Step 3: Substitute back and compute the integral

Now we have: $P(X_1 < X_2) = \int_0^{\infty} \left(1 - e^{-\theta_1 y}\right) \theta_2 e^{-\theta_2 y} \, dy.$ Expanding the terms: $P(X_1 < X_2) = \int_0^{\infty} \theta_2 e^{-\theta_2 y} \, dy - \int_0^{\infty} \theta_2 e^{-(\theta_1 + \theta_2) y} \, dy.$

1. The first integral is: $\int_0^{\infty} \theta_2 e^{-\theta_2 y} \, dy = 1.$
2. The second integral is: $\int_0^{\infty} \theta_2 e^{-(\theta_1 + \theta_2) y} \, dy = \frac{\theta_2}{\theta_1 + \theta_2}.$

So: $P(X_1 < X_2) = 1 - \frac{\theta_2}{\theta_1 + \theta_2} = \frac{\theta_1}{\theta_1 + \theta_2}.$

Conclusion

$P(X_1 < X_2) = \frac{\theta_1}{\theta_1 + \theta_2}.$ This is the required result.

Would you like further details or explanations on any part of the solution?

Related Questions

1. How does the exponential distribution affect the probability of $X_1 < X_2$?
2. How can this result be extended to more than two variables?
3. What is the intuition behind using integrals in probability for continuous random variables?
4. How would the probability change if $X_1$ and $X_2$ were not independent?
5. What other methods could be used to find $P(X_1 < X_2)$ without integrals?

Tip

For independent exponential variables $X_1$ and $X_2$, the probability $P(X_1 < X_2)$ can often be simplified using symmetry properties of the exponential distribution.