$P(T_2 \ge T_1)$ Exponentially distributed functions I have a problem regarding probability.
The lifetime $T$ of a product is exponentially distributed.
Product 1 has expected lifetime: $F_1(t) = 1-e^{-t/1000} $
Product 2 has expected lifetime: $F_2(t) = 1-e^{-t/500} $
What is the probability that product 2 outlives product one? That is:
$P(T_2 \geq T_1)$
 A: You need to know the joint distribution of $T_1$ and $T_2$ for that. I guess they are assumed to be independent in which case the probability is given by
$$ \int_0^\infty \int_0^\infty 1 \{ t_2 \geq t_1 \} f_1(t_1) f_2(t_2) dt_1 dt_2    $$
where $f_1$ and $f_2$ are the densities of the corresponding exponential distributions.
A: We have that
$$
P(T_1 \le T_2)=P(-T_2+T_1 \le 0)=F_{-T_2+T_1}(0)
$$
where $F_X$ denotes the cdf of a random variable $X.$ 
A reasonable assumption is that $-T_2$ and $T_1$ are independent. Then as a corollary of the convolution formula for the sum $-T_2+T_1$ we get that
$$
F_{-T_2+T_1}(z)=\int_0^\infty F_{-T_2}(z-t) f_{T_1}(t)dt
$$
where $f_{T_1}$ is the pdf of $F_{T_1}$ (which is your $F_1;$ details?). Since $F_1$ is continuously differentiable,
$$
f_{T_1}(t)=F_1'(t)=\frac{\mathrm{exp}(-t/1000)}{1000}.
$$ 
Now
$$
F_{-T_2+T_1}(0)=\int_0^\infty F_{-T_2}(-t) \frac{\mathrm{exp}(-t/1000)}{1000} dt
$$
and, observing that $F_{-T_2}(-t)=\mathrm{exp}(-t/500)$ (why?) we get that
$$
P(T_1 \le T_2) =\int_0^\infty \mathrm{exp}(-t/500) \frac{\mathrm{exp}(-t/1000)}{1000} dt=\frac 13.
$$
A: I'm going to guess that you intended these to be independent.  That assumption is often not mentioned but ought to be.
You can write either
$$
\int_0^\infty \left(\int_{t_1}^\infty \cdots\cdots \, dt_2 \right) \, dt_1
$$
or
$$
\int_0^\infty \left(\int_0^{t_2} \cdots\cdots \,  dt_1 \right) \, dt_2
$$
where the expression in place of "$\cdots\cdots$" is the joint density function.  Either way you're integrating over the part of the plane in which $t_2>t_1$.
If the two random variables are independent, then the joint density function is
$$\frac{1}{1000} e^{-t_1/1000}\cdot\frac{1}{500} e^{-t_2/500}$$
if $t_1,t_2\ge0$ and $0$ if either $t_1<0$ or $t_2<0$.
The first integral is slightly more convenient because you evaluate at $\infty$ and expression that approaches $0$ there.
\begin{align}
& {}\qquad \int_0^\infty \left(\int_{t_1}^\infty \frac{1}{1000} e^{-t_1/1000}\cdot\frac{1}{500} e^{-t_2/500} \, dt_2 \right) \, dt_1 \\
& = \int_0^\infty\left(\frac{1}{1000}e^{-t_t/1000} \int_{t_1}^\infty \frac{1}{500} e^{-t_2/500} \, dt_2 \right) \, dt_1 \\
& {}\qquad (\text{since the expression that was pulled out does not depend on }t_2) \\
& = \int_0^\infty \frac{1}{1000} e^{-t_1/1000} \cdot e^{-t_1/500} \, dt_1 \\
& {} \qquad (\text{The only variable still here is $t_1$ since the definite integral was evaluated.}) \\ & = \int_0^\infty \frac{1}{1000} e^{-3t_1/1000} \, dt_1 \\
& = \int_0^\infty e^{-3u}\,du = \frac 1 3.
\end{align}
A: Let's try an intuitive argument in addtion to three already posted.
One of these exponential random variables has an expected value of $1000$ and for the other it's $500$.  That means Product 2 must be replaced twice for every time Product 1 is replaced.  Therefore an average set of three expiration events consists of two expirations of Product 2 and one expiration of Product 1.  Hence the probability that the next to expire is Product 1 is $1/3$.
