Probability that all runners finish within (9.8s to 9.9s)

[psie]

Homework Statement

I guess it is Olympic games, so here's a probability question. In a $100$-meter Olympic race the running times can be considered to be $U(9.6,10.0)$-distributed. Suppose there are eight competitors in the finals. What is the probability that all eight finish within the time interval $(9.8,9.9)$?

Relevant Equations

This is in a chapter on order statistics, specifically in a section on the joint density of the extreme order statistics, i.e. $X_{(n)}=\max\{X_1,\ldots,X_n\}$ and $X_{(1)}=\min\{X_1,\ldots,X_n\}$. Their joint density is $$f_{X_{(1)},X_{(n)}}(x,y)=\begin{cases} n(n-1)(F(y)-F(x))^{n-2}f(y)f(x),&x<y\\ 0&\text{otherwise}.\end{cases}$$ The range is $R_n=X_{(n)}-X_{(1)}$ and it also has a known density which may be of interest (not sure).

I'm not sure how to solve this. Intuitively, we'd want to know $$P(X_{(1)}>9.8,X_{(8)}<9.9),$$but is the above probability simply $P((X_{(1)},X_{(8)})\in (9.8,10.0)\times(9.6,9.9))$ and do I integrate the joint density over $(9.8,10.0)\times(9.6,9.9)$ then?

 

[ergospherical]

Solve it for some small $n$ first. What's the probability for just $n=1$ runner? And what about for $n=2$ runners? In the latter case, you can think in terms of the joint probability being proportional to the area of a particular square in the $(X_1, X_2)$ plane. And for $n=3$ runners, in terms of a particular cubic volume in $(X_1, X_2, X_3)$ space. Can you generalise?

[FYI, might be worth ditching the decimals and doing a simple re-scaling to consider the running times to be $U(0,4)$ distributed, and consider positive results to be those in the interval $(2,3)$. Just easier to deal with and/or sketch...]

 

[Orodruin]

FWIW, this problem is much much simpler to solve using that the running times are independent and all need to fall within a particular interval. The probability of all runners ending up in that interval is simply the product of the probabilities for each runner ending in the target interval. Don't overcomplicate things just because it is in a particular chapter.

* Also, let me strongly doubt this model of olympic running times ...

 

[psie]

Here's my solution. The range has density $$f_{R_n}(r)=n(n-1)\int_{-\infty}^\infty (F(u+r)-F(u))^{n-2}f(u+r)f(u)\,du,$$for $r>0$. We want to know the probability $$P(R_8\leq 0.1)=\int_0^{0.1}f_{R_8}(r)\,dr.$$To compute this probability, we need to compute $f_{R_8}(r)$. But also $f$ and $F$, which are luckily quite simple; $f(x)=\frac1{10-9.6}=2.5$ for $9.6<x<10$ and $0$ otherwise, and $F(x)=2.5x-24$ for $9.6<x<10$, $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>1$.

Now we can compute $f_{R_8}(r)$. We notice first that the integral is zero except when $9.6<u<10-r$, so: \begin{align*} f_{R_8}(r)&=56\int_{9.6}^{10-r}(2.5(u+r)-24-2.5u+24)^6 2.5^2\,du\\ &=56\cdot(2.5)^8 r^6\int_{9.6}^{10-r}du \\ &=56\cdot(2.5)^8 r^6(0.4-r).\end{align*} Integrating $f_{R_8}$ from $0$ to $0.1$ gives $0.00038$ as WolframAlpha confirms.

Now, this does not agree with \begin{align*}P(X_{(1)}>9.8,X_{(8)}<9.9)&=P(X_1\in (9.8,9.9),\ldots,X_8\in (9.8,9.9))\\ &=P(X_1\in (9.8,9.9))^8 \\ &=\frac14^8\approx 0.000015\end{align*}Where did I go wrong?

 

[Orodruin]

By computing the integral of the range density function, you seem to be computing the probability that the range is less than 0.1, but this is not the sought probability. There are outcomes that satisfy this but do not satisfy the required condition of all runners in the interval (9.8,9.9), such as all runners running exactly 9.7 (which would make range zero but not satisfy all runners in the required interval).

 

[Orodruin]

In a stroke of coincidence, the Olympic final saw all 8 runners ending up with times in the interval (9.79,9.91). Close enough?

I’m going to call p<0.05 on ruling out the suggested distribution of times at 95% CL

Edit: This is what a tight race looks like:

 

[Vanadium 50]

In a stroke of coincidence

Sure they didn't do this just for you?