Streaky Shooter, Stephen Curry

So, recently I was given the following problem: let's say some basketball player has an a priori probability  $P(y_1=1)=0.35$ of making  a three point shot. This implies the probability of missing the three-pointer is $P(y=0)=0.65$. (Note that $y=1$ denotes ``make'' and $y=0$ ``miss''.) Pretty good odds, for an NBA player -- I think.

But let's say that our basketball player is streaky, in the sense that $P(y_2=1 \vert y_1=1)=0.6$, or the probability of making the next shot, given that he made the previous, is sixty percent. (This implies $P(y_2=0 \vert y_1=1)=0.4$.) And conversely, the probability of making the next shot, given that he missed the previous is a bit lower than the prior, $P(y_2=1 \vert y_0=0)=0.3$, which implies $P(y_2=0 \vert y_0=0)=0.7$. This defines something called a markov process, in the sense that the $n^\mathrm{th}$ shot only depends on the one before it.

\begin{eqnarray}
P(y_n \vert y_{n-1},y_{n-2},\dots y_2,y_1) &=& P(y_n \vert y_{n-1})
\end{eqnarray}

To repeat an old aphorism: ``the future to depends on the past only through the present''. We can represent this conditional probability density with a matrix,

\begin{eqnarray}
P(y_n \vert y_{n-1}) &=& \mathbf{W}= \left(\begin{array}{cc}
.6 & .3 \\
.4 & .7 \\
\end{array}\right)
\end{eqnarray}

and the a priori probability density with a vector:

\begin{eqnarray}
P(y_1) &=& \left(\begin{array}{c}
0.35 \\
0.65
\end{array}\right)
\end{eqnarray}

If we want to write down the joint probability density on a few shots, we can use Bayes' chain rule:

\begin{eqnarray}
P(y_3, y_2,y_1) &=& P(y_3 \vert y_2, y_1) P(y_2 \vert y_1) P(y_1)
\end{eqnarray}

which by markov property simplifies to

\begin{eqnarray}
P(y_3, y_2,y_1) &=& P(y_3 \vert y_2) P(y_2 \vert y_1) P(y_1)
\end{eqnarray}

Marginalizing over $y_2$ we see that we have a matrix multiplication:

\begin{eqnarray}
\sum_{y_2} P(y_3, y_2,y_1) &=& \sum_{y_2}  P(y_3 \vert y_2) P(y_2 \vert y_1) P(y_1)\\
P(y_3,y_1) &=& \sum_{y_2}  P(y_3 \vert y_2) P(y_2 \vert y_1) P(y_1)
\end{eqnarray}

and, dividing through by $P(y_1)$ is the same as conditioning on $y_1$, so we have a new conditional probability, which connects the third shot to the first:

\begin{eqnarray}
P(y_3 \vert y_1) &=& \sum_{y_2}  P(y_3 \vert y_2) P(y_2 \vert y_1)
\end{eqnarray}

This is often called the Chapman Kolgomorov equation, and is the basis for much analysis in physics of the rate of change of probability distributions through the master equation and Fokker-Planck equation.

It's pretty easy to see that any conditional distribution can now be written as $n-1$ matrix contractions:

\begin{eqnarray}
P(y_n \vert y_1) &=& (\mathbf{W})^{n-1}
\end{eqnarray}

If we multiply by the vector $P(y_1)$, we get the distribution on $y_n$ alone, since

\begin{eqnarray}
P(y_2) &=& \sum_{y_1} P(y_2 \vert y_1)P(y_1)\\
&=& \mathbf{W}\cdot \vec{\mathbf{P}}_1 \\
P(y_n) &=& (\mathbf{W})^{n-1} \cdot \vec{\mathbf{P}}_1
\end{eqnarray}

But Markov matrices, since their columns add to unity and have strictly non-negative entries have some very useful properties (the Perron-Frobenius Theorem). Their spectra -- or eigenvalues -- all lie within the unit circle in the complex plane, and there is guaranteed to be some eigenvector $\mathbf{\pi}$ with an eigenvalue $\lambda=1$. This eigenvector is called the stationary distribution, and after $n \to \infty$ time steps, we expect our distribution to converge to $\mathbf{\pi}$. For the matrix given above, this vector is

\begin{eqnarray}
\mathbf{\pi} &=& \left(\begin{array}{c}
0.428 \\
0.572
\end{array}\right)
\end{eqnarray}

So after sufficient shots, our shooter will converge to a 42.8 percent success rate, as compared to 35 -- or whatever you specified. This discrepancy is important, as the expected number of buckets given $m$ shots does not go like a binomial distribution

\begin{eqnarray}
\langle \sum_i y_i \rangle &=& m p_0 \\
p_0=0.35
\end{eqnarray}

but something slightly different. More like a binomial process with changing parameter $p$. As $m$ gets large we expect

\begin{eqnarray}
\langle \sum_i y_i \rangle &\approx & m \pi_0 \\
\pi_0 &=& 0.428 = \frac{3}{7}
\end{eqnarray}