I have some elementary queries about the relationship between exchangeability and conditional distributions.

For theorem 11.6.4 in the Doctrine of Chances, Ethier S. (2010) it is proved that $\{Z_n\}_{n=0, \dots, N-m}$ is a (Doob) martingale with respect to $\{X_n\}_{n=1, \dots N-m}$, where:

$Y_n = f(X_{n+1}, \dots, X_{n+m})$ where $f$ is some deterministic function that is also bounded.

$Z_n = \mathbb{E} [Y_n \vert X_1, \dots, X_n]$

And where $N = 52$, $m > 0$, and $X_1, \dots, X_N$ are exchangeable.

It can be shown that $\mathbb{E}[\vert Z_n \vert] < \infty$ from boundedness of $f$. That there exist functions such that $Z_n = h_n(X_1, \dots, X_n)$ arises from the definition of conditional expectation. This satisfies the subsidiary conditions of a (Doob) martingale.

The demonstration of the final condition that $Z_n = \mathbb{E}[Z_{n+1} \vert X_1 \dots, X_n ]$ uses the generalised law of iterated expectations and exchangeability:

$$\begin{align} \mathbb{E}[Z_{n+1} \vert X_1, \dots, X_n] &= \mathbb{E} \left[\mathbb{E}[Y_{n+1} \vert X_1, \dots X_{n+1}] \vert X_1 \dots X_n \right] \\ &= \mathbb{E}\left[Y_{n+1} \vert X_1 \dots, X_n \right] \\ &= \mathbb{E}\left[Y_{j} \vert X_1 \dots, X_n \right] \quad \quad \forall \space j \geq n \\ \end{align}$$

Taking $j = n$, the final condition for this to be a martingale that $\mathbb{E}[Z_{n+1} \vert X_1, \dots, X_n] = Z_n$ is satisfied.


The crux of the argument is that $Y_{n+1}$ and $Y_j$ have the same conditional expectation. Is that because they have the same conditional distribution, arising from exchangeability? And if so, would the following attempt to show how this arises from the definition of exchangeability be correct?

We want to show that for some $j \geq n$:

$$\mathbb{E}\left[Y_{n+1} \vert X_1 \dots, X_n \right] = \mathbb{E}\left[Y_{j} \vert X_1 \dots, X_n \right]$$

Where $Y_{n+1} = f(X_{n+2}, \dots X_{n+1+m})$ and $Y_j = f(X_{j+1}, \dots X_{j+m})$

Exchangeability of a sequence of random variables means that their joint distribution is invariant to any permutations $\sigma$ of the indices of the random variables:

$$p(X_1, \dots, X_N) \overset{d}{=} p(X_{\sigma(1)}, \dots, X_{\sigma(N)})$$

We can rewrite this as

$$p(X_{n+2}, \dots, X_{n+1+m} \vert X_1, \dots, X_n) p(X_1, \dots, X_n) = p(X_{j+1}, \dots, X_{j+m} \vert X_1, \dots, X_n) p(X_1, \dots, X_n)$$

And so

$$p(X_{n+2}, \dots, X_{n+1+m} \vert X_1, \dots, X_n) = p(X_{j+1}, \dots, X_{j+m} \vert X_1, \dots, X_n)$$

Further context.

As context was not required for the question, I have placed it here. The proof that $\{Z_n\}$ is a (Doob) martingale is a stepping stone to showing that using information about the cards have been dealt in Blackjack allows profit opportunities in the form of an increasing $\text{Var}[Z_n]$ as the deck is progressively depleted.

  • $N = 52$ refers to the number of cards in a 52 card deck.
  • $X_1, \dots, X_N$ are the card labels in a 52 card deck.
  • $f$ is a deterministic function of the $m$ cards dealt in a coup that gives the player's profit per unit bet whilst playing according to a decision rule known as 'Basic Strategy'.
  • $m > 0$ is the number of cards required to complete a coup with the dealer playing according to House rules and a player playing Basic Strategy.
  • $Y_n$ is the player profit per unit bet
  • $Z_n$ is the conditional expected profit per unit bet after having seen $n$ cards $X_1, \dots X_n$.

1 Answer 1


Firstly, yes, you are correct that the crux of this step is that $Y_{n+1}$ and $Y_j$ ($j \geqslant n$) have the same conditional expectation, which arises from the fact that they have the same conditional distribution. For simplicity, let's consider the case where $j>n$ (though the case $j=n$ is also simple). Consider the vector:

$$\underbrace{X_1,...,X_n}_\text{Conditioning}, \underbrace{X_{n+2}, ..., X_{n+m+1}}_{\text{Determines } Y_{n+1}}, X_{n+1}, X_{n+m+2}, ..., X_{j+m}.$$

We can permute this vector to get the alternative:

$$\underbrace{X_1,...,X_n}_\text{Conditioning}, \underbrace{X_{j+1}, X_{n+2}, ..., X_{j+m}}_{\text{Determines } Y_j}, X_{n+1}, ..., X_{j}.$$

Since the sequence $\{ X_i \}$ is exchangeable, we therefore have equivalence of the joint distributions:

$$p(X_1,...,X_n,Y_{n+1}) = p(X_1,...,X_n,Y_j),$$

which means we also get equivalence of the conditional distributions:

$$p(Y_{n+1} | X_1,...,X_n) = p(Y_j | X_1,...,X_n).$$

  • $\begingroup$ Thanks as always Ben! I was feeling really stuck with this one. $\endgroup$
    – microhaus
    Commented Jun 12, 2023 at 5:54

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