# Combining two exponential distributions onto one graph

In a probability question I am working through, the question says the length of time of a football game, $s$ , is exponentially distributed with a mean of $\bar s$ . It then says to write out the probability distribution for $s$, which I can do. The time after which the game is cancelled due to bad weather, $t$, is also exponentially distributed with a mean of $\bar t$, and $t$ is independent of $s$ .

It then asks ''Draw a pair of axes at right-angles to each other labeling one $s$ and one $t$. Indicate on the diagram the region where the game is not interrupted by bad weather.

I am confused by this, as surely I need a third axis orthogonal to the $st$ plane in order to sketch the probability distribution for both of these distributions. If I have $t$ as the independent variable, and s as the dependent variable, I can sketch $t$ as just a general exponential distribution, but then what does the sketch of $s$ against $t$ look like ?

• I think you're supposed to draw the graph of each on the same Cartesian plane. Think about graphing $f(x)=\frac{1}{x^2}$ and $g(x)=\sqrt{x}$ on the same plane. Jan 8 '16 at 10:57
• But if the graph is of $t$ against $s$ , then as $t$ changes we are looking at how $s$ changes with it, not $f(t)$ . Jan 8 '16 at 11:01
• No you are not supposed to draw two PDFs on the same diagram, actually you are not even supposed to draw any PDF. In the $(s,t)$-quarter of plane $s>0$, $t>0$, you are asked to determine the $(s,t)$-points such that the game is not interrupted by bad weather. Obviously this corresponds to the angular sector $0<s<t$, question solved.
– Did
Jan 8 '16 at 12:11
• @Did Why not post your solution as an answer instead of as a comment? Jan 8 '16 at 12:26
• @Math1000 Please feel free to do so.
– Did
Jan 8 '16 at 12:27

Let $\lambda = \frac1{\bar s}$ and $\mu=\frac1{\bar t}$, and $E$ denote the event "the game is not interrupted by bad weather". $E$ is simply the set $\{S<T\}$, and it is straightforward to compute this probability (assuming $S$ and $T$ are independent). Let $f_S$ and $f_T$ be the probability density functions of $S$ and $T$ respectively, then the joint density is equal to the product of the marginal densities, i.e. $f_{S,T}=f_Sf_T$. Hence \begin{align} \mathbb P(S<T) &= \iint\limits_{\{(s,t)\in\mathbb R^2\ :\ s<t\}} f_{S,T}\ \mathsf d(s\times t)\\ &= \int_0^\infty \int_0^t f_S(s)f_T(t)\ \mathsf ds\ \mathsf dt\\ &= \int_0^\infty \mu e^{-\mu t} \int_0^t \lambda e^{-\lambda s}\ \mathsf ds\ \mathsf dt\\ &= \int_0^\infty \mu e^{-\mu t}(1-e^{-\lambda t})\ \mathsf dt\\ &= \int_0^\infty \mu e^{-\mu t}\mathsf dt - \mu\int_0^\infty e^{-(\lambda+\mu) t}\ \mathsf dt\\ &= 1 - \frac\mu{\lambda+\mu}\\ &= \frac\lambda{\lambda+\mu}. \end{align} By symmetry it is clear that $\mathbb P(S<T)=\frac12$ if $\lambda=\mu$, in which case your plot would simply be of the set $$\{(s,t)\in\mathbb R^2: 0<s<t\}.$$ In general, I believe you are being asked to plot the set $$\left\{(s,t)\in\mathbb R^2: 0 < s < \frac\mu\lambda t \right\}.$$ My intuition is that for a fixed $\mu$, we have \begin{align} \lim_{\lambda\to0}\frac\lambda{\lambda+\mu}&=0\\ \lim_{\lambda\to\infty}\frac\lambda{\lambda+\mu}&=\infty, \end{align} and the plot with the slope $\frac\mu\lambda$ is a function that comes to mind that satisfies that property.
• @Math1000 , the next part of the question is actually a double integral to find the probability of $0<S<T$ over the region Did answered, so this is still a useful answer:)thanks Jan 8 '16 at 15:13