A Coin with Random Bias

What this lecture covers

This is a worked recitation problem that layers randomness on top of randomness: a coin is flipped n times, but the coin's own bias is a random variable Q with known mean mu and variance sigma squared. The problem asks for the expectation of individual flips, the covariance between flips, and the variance of the total number of heads, computed two different ways.

It brings together several tools from the main lecture on iterated expectations: the law of iterated expectations, the covariance formula, and the law of total variance, applied to a hierarchy where the randomness of each coin flip depends on the randomness of the bias above it. After watching, you should be able to recognize when conditioning on a hidden parameter simplifies a layered random experiment, and compute covariance between dependent Bernoulli trials that share a common random bias.

Key ideas

• Setup: Xi is 1 if the i-th flip is heads, 0 otherwise; conditional on Q, all flips are independent; E[Q] = mu, Var(Q) = sigma^2 > 0.
• Conditional expectation of a flip: E[Xi|Q] = Q, so by the law of iterated expectations E[Xi] = E[Q] = mu, and E[X] = n*mu for the total number of heads X.
• Covariance between different flips is not zero: for i != j, Cov(Xi, Xj) = E[Q^2] - mu^2 = Var(Q) = sigma^2, because sharing an unknown bias correlates the flips even though they are conditionally independent given Q.
• Correlated implies not independent: since sigma squared is positive, Xi and Xj are correlated, and correlated random variables cannot be independent (though the converse - uncorrelated implies independent - is not generally true).
• Variance of a single flip: Var(Xi) = mu - mu^2.
• Two ways to compute Var(X): the law of total variance (conditioning on Q) and the covariance-sum formula (summing variances plus all pairwise covariances) give the same result, n*(mu - mu^2) + n*(n-1)*sigma^2.
• Rule of thumb for layered randomness: when one layer of randomness (coin flips) depends on another (the bias), condition on the upper layer to simplify calculations at the lower layer.

Walkthrough

Setting up notation and conditional independence (0:00)

The problem defines the random bias Q, the Bernoulli indicators Xi for each flip, and states the key assumption that flips are independent conditional on Q, which will be used throughout.

Expectation of a single flip and of the total (1:01)

Conditioning on Q gives E[Xi|Q] = Q, so by the law of iterated expectations E[Xi] = mu; summing over n flips using linearity of expectation gives E[X] = n*mu.

Covariance between flips (5:06)

For i != j, the problem computes E[Xi*Xj] by conditioning on Q and using conditional independence, arriving at Cov(Xi, Xj) = sigma^2, showing the flips are correlated because they share the same unknown bias. It also computes Var(Xi) = mu - mu^2 for the i = j case, using the fact that a Bernoulli variable squared equals itself.

Variance of the total via the law of total variance (12:22)

Conditioning on Q, the problem computes E[Var(X|Q)] = n*(mu - sigma^2 - mu^2) and Var(E[X|Q]) = n^2 * sigma^2, combining them into Var(X) = n*(n-1)*sigma^2 + n*(mu - mu^2).

Cross-check via the covariance-sum formula (18:45)

The problem recomputes Var(X) by summing the n individual variances and the n^2 - n pairwise covariances directly, confirming it matches the law-of-total-variance result, and closes with a rule of thumb for choosing what to condition on in layered random experiments.

Before you watch

• Watch the main lecture on iterated expectations for the law of iterated expectations, the covariance formula, and the law of total variance.
• Know the mean and variance of a Bernoulli random variable.

Check your understanding

1. Why are Xi and Xj correlated even though they are conditionally independent given Q?
2. Derive Cov(Xi, Xj) = sigma^2 for i != j.
3. Walk through both methods used to compute Var(X) and confirm they agree.
4. What rule of thumb does the problem suggest for choosing what to condition on in a layered random experiment?