A theoretical look at stochastic gradient descent

Fall 2022: Peter Ralph

https://uodsci.github.io/dsci345

Application: stochastic gradient descent

Problem: find the mean

Suppose we have a very large dataset, so large (or unwieldy) that we don't get to see all of it at once. Instead, we get fed it in randomly chosen batches, each of size $n$. We would like to do something simple: estimate the "typical value".

Question: How would you do this?

Fact: For any collection of numbers $x_1, \ldots, x_N$, the mean is the unique value that minimizes the mean-squared error, i.e., if $\mu = (1/N) \sum_{i=1}^N x_i$ and $$ F_x(m) = \frac{1}{N} \sum_{i=1}^N (x_i - m)^2 ,$$ then $\min_m F_x(m) = F_x(\mu)$.

Proof: The derivative of $F(m)$ is $$ F_x'(m) = 2 \frac{1}{N} \sum_{i=1}^N (m - x_i) = 2(m - \mu) .$$ This is equal to zero only at $m = \mu$.

Minimization

Here's a solution: first, start with a guess $m = 0$, and pick an $\epsilon > 0$. Then, for each batch $X_1, \ldots, X_n$:

1. Compute the batch mean $\bar X = (1/n) \sum_{i=1}^n X_i$.

2. Compute the gradient of $F$ using this batch, $F_X'(m) = 2 (m - \bar X)$.

3. If $F_X'(m)$ is small, then stop. Otherwise, move our guess $m$ a little bit, updating $m \mapsto m - \epsilon F_X'(m)$.

How's it work in practice?

# set-up
mu = 3
n = 10

def next_batch():
    return rng.exponential(scale=mu, size=n)

m = 0
eps = 0.1  #  play around with this!
num_batches = 100
mvec = np.repeat(np.nan, num_batches)
mvec[0] = m
for k in range(1, num_batches):
    X = next_batch()
    dF = 2 * (m - np.mean(X))
    m -= eps * dF
    mvec[k] = m

plt.plot(mvec)
plt.xlabel("batch number"); plt.ylabel("estimate")
plt.title(f"espilon = {eps}")
plt.hlines(mu, 0, num_batches, "red", ":"); 

That's nice, but what about theory?

Well, we know that if $X_1, \ldots, X_n$ are independent and identically distributed (iid) with mean $\E[X_i] = \mu$ and standard deviation $\text{sd}(X_i) = \sigma$, then $$ \E[\bar X] = \mu $$ and $$ \text{sd}[ \bar X ] = \frac{\sigma}{\sqrt{n}} . $$

So, the expected change is $$ \E[F'_X(m)] = \E[2 (m - \bar X)] = 2 (m - \mu) , $$ and the standard deviation is $$ \text{sd}[F'_X(m)] = \frac{2 \sigma}{\sqrt{n}} . $$

Conclusion

• If $m = \mu$ then there is no mean change, i.e., "equilibrium" is at $m = \mu$.
• If $|m - \mu| \gg 2\sigma/\sqrt{n}$, then the mean change is more important than noise.
• If $|m - \mu| \le 2\sigma/\sqrt{n}$ then the "noise" is important.

So, this will fluctuate around $\mu$, varying by an amount proportional to $\sigma/\sqrt{n}$. (For the exponential, $\sigma=\mu$.)

Doodling in office hours

What happens if the mean changes with time? Let's have it switch back and forth between 2 and 4 every 200 steps.

# set-up
n = 10

def next_batch(t):
    if int(t/200) % 2 == 0:
        mu = 2
    else:
        mu = 4
    return rng.exponential(scale=mu, size=n)

m = 0
eps = 0.05  #  play around with this!
num_batches = 1000
mvec = np.repeat(np.nan, num_batches)
mvec[0] = m
for k in range(1, num_batches):
    X = next_batch(k)
    dF = 2 * (m - np.mean(X))
    m -= eps * dF
    mvec[k] = m

plt.plot(mvec)
plt.xlabel("batch number"); plt.ylabel("estimate")
plt.title(f"espilon = {eps}")
plt.hlines([2, 4], 0, num_batches, "red", ":");