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So this is a question from the Rice textbook on mathematical statistics. It is as follows:

Let $T_1$, $T_2$ be independent Exponential random variables with parameters $\alpha$ and $\beta$. Calculate the probability that $T_1 < T_2$.

So I am a bit confused as to how I should start this question. After some research online and consultation with friends, I was told to use the joint distribution of the two functions (which is easy to calculate as the R.Vs are independent and thus we can just multiply their respective probability distributions). However, after this a wacky integral is set up, usually of the form: $$\int_{0}^{\infty}\int_{0}^{y} f(x,y)dx dy$$ What does this mean? Intuitively speaking, is this the integral of the joint density under the line $y = x$? But these are random variables, ($T_1$,$T_2$), so how would I set up the integral in this case? I am just very confused overall with regard to joint distributions and am not entirely sure where to start.

rubikscube09
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2 Answers2

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$$\color{blue}{\int_0^\infty} \color{red}{\int_0^y} f_{T_1, T_2}(x,y) \,\color{red}{dx}\,\color{blue}{dy}$$

$f(x,y)$ is the density function, upon integrating them, we recover the corresponding probability of the interested event.

The integration domain describe the region of interest.

Here $x$ correponds to $T_1$ and $y$ correponds to $T_2$.

The domain of integration does describe the set $\{(x,y)|y \geq x, y\geq 0 , x\geq 0\}$.

First, $y$ can take any nonnegative value. Upon fixing $y$, $x$ can only take value from $0$ up to $y$.

Siong Thye Goh
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You are correct with the integral set up. Essentially, you are looking at the event $[T_1 = t]\ \cap [T_2 \geq= t]$, for all possible values of $t$. So your inner integral is the probability that $X$ is less than some fixed value $y$ (i.e. between $0$ and $y$, and you integrate over all possible values that $Y$ can take.

In this particular case, with independence, your equation should look like this:

$$\int_{0}^{\infty}\int_{0}^{t_2} f_{T_1}(t_1) \ f_{T_2}(t_2) \ dt_1 dt_2 = \int_{0}^{\infty}F_{T_1}(t_2)\ f_{T_2}(t_2) \ dt_2 = \mathbb{E}_{T_2}\left[F_{T_1}(T_2)\right]$$

Ken Wei
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  • Right, so $t$ is some fixed value corresponding to $T_2$? And then fixing this value of $t$ we evaluate the marginal distribution/CDF of $T_1$. This is what I think at least the inner integral can be interpreted as. Then we let $t$ vary over all possible values of $T_2$. – rubikscube09 Oct 30 '17 at 06:39
  • Yes, that's exactly it. – Ken Wei Oct 30 '17 at 06:41