Will You Top the Leaderboard?

This Week's Fiddler

We note that for each rider, there are two uniform distributions at play here: power ($P=U(0,200)$, in Watts) and time ($T=U(0,1800)$, in seconds). The product of these, $W=PT$, represents the work this rider has done, which is what the leaderboard ranks by.

If we're at the middle of the leaderboard, this means our work $w$ rests at the median of $W$. Now we need to determine $F_W(w)$, the cumulative distribution function of $W$.

$$\begin{align*} F_W(w)&=\Pr(W\le w) \\[1em] &=\frac{1}{200}\int_0^{200}\Pr\Bigl(T\le\frac{w}{p}\Bigr)\textrm{ d}p \\[1em] &=\frac{1}{200}\Biggl(\int_0^{\frac{w}{1800}}\textrm{ d}p+\int_{\frac{w}{1800}}^{200}\Pr\Bigl(T\le\frac{w}{1800\cdot p}\Bigr)\textrm{ d}p\Biggr) \\[1em] &=\frac{1}{200}\Biggl(\biggl[p\biggr]_0^{\frac{w}{1800}}+\biggl[\frac{w}{1800}\ln(p)\biggr]_{\frac{w}{1800}}^{200}\Biggr) \\[1em] &=\frac{1}{200}\Biggl(\frac{w}{1800}+\frac{w}{1800}\cdot\ln(200)-\frac{w}{1800}\ln\Bigl(\frac{w}{1800}\Bigr)\Biggr) \\[1em] &=\frac{w}{360000}\biggl(1-\log\Bigl(\frac{w}{360000}\Bigr)\biggr) \end{align*}$$

Now to find the median work $w_m$:

$$0.5=F_W(w_m)=\frac{w_m}{360000}\Bigl(1-\log(\frac{w_m}{360000})\Bigr)$$

From this, we find that $w_m\approx67205.6$.

This is our work in the middle of our workout. But what about the end? Well, assuming we keep pace, it should be double this at $134411.2\textrm{ J}$. Now we want to find the rank of this work:

$$F_W(134411.2)=\frac{134411.2}{360000}\biggl(1-\log\Bigl(\frac{134411.2}{360000}\Bigr)\biggr)\approx0.7412$$

So we can expect to be $74.12\%$ up the leaderboard (top $25.88\%$).

As for the bonus question, the highest work we can expect to be at the 15-minute mark of our workout is $200\textrm{ W}\cdot900\textrm{ s}=180000\textrm{ J}$. Hence, we can find its rank:

$$F_W(180000)=\frac{180000}{360000}\biggl(1-\log\Bigl(\frac{180000}{360000}\Bigr)\biggr)\approx0.8466$$

That would be about $84.66\%$ up the leaderboard (top $15.34\%$).

Answer: 74.12% up the leaderboard (bonus: 84.66% up the leaderboard)

Extra Credit

To find our expected rank up the leaderboard given a random power and random time, we can find weighted $f_W$ over the integral from $p=0$ to $p=200$ and $t=0$ to $t=1800$.

$f_W(w)=-\frac{1}{360000}\log(\frac{w}{360000})$

The weighted rank at each point is $F_W(w)$. So our expectation is:

$$\int_0^{1800} \int_0^{200} f_W(p\cdot t)\cdot F_W(p\cdot t)\textrm{ d}p\textrm{ d}t \\ \int_0^{1800} \int_0^{200} \biggl(\frac{-\log\bigl(\frac{p\cdot t}{360000}\bigr)}{360000}\biggr)\biggl(\frac{p\cdot t}{360000}\biggr)\biggl(1-\log\Bigl(\frac{p\cdot t}{360000}\Bigr)\biggr)\textrm{ d}p\textrm{ d}t$$

A gnarly-looking integral, but it resolves cleanly to $\frac{5}{8}$, or $62.5\%$ up the leaderboard.

Answer: 62.5% up the leaderboard