Probability of $APreparing exams, I ran into the following problem:


Edit: it shouldn't be represented as it was. Added the storkes.

Let $A$, $B$ be two independent variables having probability distribution:
$$
\require{cancel}
\cancel{P(A = a) =} f_a(t) = \begin{cases} \lambda_ae^{-\lambda_a t} &, x \geq 0 \\ 0 &, x < 0 \end{cases}\\
\cancel{P(B = b) =} f_b(t) = \begin{cases} \lambda_be^{-\lambda_b t} &, x \geq 0 \\ 0 &, x < 0 \end{cases}
$$
Or in short, having exponential distribution with parameter $\lambda_a$ and $\lambda_b$, respectively.
What is the probability of $A < B$?

What I thought to do is to do the following integral since the value of possible $A$ is bounded by $B$:
$$
\displaystyle \int^{\infty}_{0} \left( \int^{f_b(t)}_{0} f_a(t) dt\right)dt = \int^{\infty}_0 e^{-\lambda_a\lambda_b e^{-\lambda_b t}} dt
$$
which produces an integral using the exponential integral function $E_1(x)$, which is very weird as an exam problem. But where did I think wrong and how should I approach this problem?

You integrate the joint distribution $f_A(a)f_B(b)$ over the region $a<b$
$$P(A<B) = \int^{\infty}_{0}  \int^{\color{blue}{b}}_{0} f_A(a)f_B(b) \,\color{red}{da} \,\color{red}{db}$$

In your question you have several issues

*

*You are referring to probabilities like $P(A=a)$, but these are zero. Your function $f_A$ is a density function, and that's not the same as a probability.


*Your integral uses $dt$ twice


*You have the density as limit of the integrand instead of the variable
These last two points I have coloured in the function above. You can compare it to your function below where I coloured it the same.
$$\displaystyle \int^{\infty}_{0} \left( \int^{\color{blue}{f_b(t)}}_{0} f_a(t) \color{red}{dt}\right)\color{red}{dt}$$
