# The Information Structuralist

## Blackwell’s proof of Wald’s identity

Posted in Mathematics, Probability by mraginsky on April 29, 2011

Every once in a while you come across a mathematical argument of such incredible beauty that you feel compelled to tell the whole world about it. This post is about one such gem: David Blackwell’s 1946 proof of Wald’s identity on the expected value of a randomly stopped random walk. In fact, even forty years after the publication of that paper, in a conversation with Morris DeGroot, Blackwell said: “That’s a paper I’m still very proud of. It just gives me pleasant feelings every time I think about it.”

Let ${X^\infty = (X_1,X_2,\ldots)}$ be a sequence of independent real-valued random variables with common mean ${\mu}$. Let ${\tau}$ be a stopping time, i.e., a random variable taking values in ${{\mathbb N}}$, such that the occurrence or nonoccurrence of the event ${\{\tau=n\}}$ is determined only by ${X^n := (X_1,\ldots,X_n)}$. We say that Wald’s identity holds for the pair ${(X^\infty,\tau)}$ if $\displaystyle \mathop{\mathbb E}\left[ X_1 + \ldots + X_\tau\right] = \mu \cdot \mathop{\mathbb E} \tau.$

The first result of this type has appeared in the seminal paper of Abraham Wald on sequential hypothesis testing (hence the name), and it can be stated as follows:

Theorem 1 Assume that ${X_1,X_2,\ldots}$ are i.i.d., ${\mathop{\mathbb E} |X_i| = \mathop{\mathbb E} |X_1| < \infty}$, and ${\mathop{\mathbb E} \tau < \infty}$. Then Wald’s identity holds for the pair ${(X^\infty,\tau)}$.

The standard proof of this theorem uses martingale arguments. In 1946, David Blackwell found a beautiful proof that uses only Kolmogorov‘s Strong Law of Large Numbers (SLLN) and its converse. Since the latter is, perhaps, less well known than the former, I will just give both for convenience (see, e.g., Chapter 12 in Probability With Martingales by David Williams):

1. SLLN: Let ${U_1,U_2,\ldots}$ be i.i.d. with ${\mathop{\mathbb E} |U_i| = \mathop{\mathbb E} |U_1| < \infty}$ for all ${i}$. Then $\displaystyle \lim_{n \rightarrow \infty} \frac{U_1 + \ldots + U_n}{n} = \mathop{\mathbb E} U_1 \qquad \text{a.s.}$

2. Converse to SLLN: If ${U_1,U_2,\ldots}$ are i.i.d. with ${\mathop{\mathbb E} |U_i| = \mathop{\mathbb E} |U_1| = \infty}$, then $\displaystyle \lim_{n \rightarrow \infty} \frac{U_1 + \ldots + U_n}{n} \text{ does not exist a.s.}$

Blackwell’s proof

Let us define recursively a sequence of random variables ${\tau_1,\tau_2,\ldots}$ as follows. Let ${\tau_1 = \tau}$. Then, assuming ${\tau_1,\ldots,\tau_k}$ have been defined, let $\displaystyle \tau_{k+1} = \tau(X_{\tau_1+\ldots+\tau_k+1}, X_{\tau_1 + \ldots + \tau_k + 2},\ldots).$

In other words, ${\tau_1}$ is the stopping time determined by the original full sequence ${X^\infty}$, ${\tau_2}$ is the stopping time determined by the remaining data after the first stopping time, i.e., by ${X^\infty_{\tau_1+1}}$, ${\tau_3}$ is the stopping time determined by the data remaining after the second stopping time, i.e., by ${X^\infty_{\tau_1+\tau_2 + 1}}$, etc. We now claim that ${\tau^\infty = (\tau_1,\tau_2,\ldots)}$ is an i.i.d. sequence. To see this, let us compute the conditional probability that ${\tau_{k+1} = n}$ given ${\tau^k = (\tau_1,\ldots,\tau_k)}$. Given positive integers ${j_1,\ldots,j_k}$, define the event $\displaystyle S = \left\{ \tau_1 = j_1,\ldots,\tau_k = j_k \right\}$

as well as the event $\displaystyle R = \left\{ \tau(X^\infty_{j_1+\ldots +j_k +1}) = n \right\}.$

By construction, the occurrence or nonoccurrence of ${S}$ is determined only by ${X^{j_1 + \ldots + j_k}}$, so ${S}$ is independent of ${R}$. Therefore, $\displaystyle \begin{array}{rcl} \Pr\Big(\tau_{k+1}= n |\tau_1 = j_1,\ldots,\tau_k = j_k\Big) &=& \frac{\Pr\left(S \cap \left\{ \tau_{k+1} = n\right\}\right)}{\Pr(S)} \\ &=& \frac{\Pr(S \cap R)}{\Pr(S)} \\ &=& \Pr(R). \end{array}$

Moreover, since the ${X_i}$‘s are i.i.d., ${\Pr(R) = \Pr(\tau_1 = n) =: \pi_n}$. Thus, we have shown that, for every ${k = 1,2,\ldots}$ and every ${n}$, $\displaystyle \Pr(\tau_{k+1} = n |\tau_1,\ldots,\tau_k) = \pi_n,$

from which we conclude that ${\tau_1,\tau_2,\ldots}$ is an i.i.d. sequence.

Now let us define, for each ${k}$, the random variable $\displaystyle S_k = X_{\tau_1 + \ldots + \tau_{k-1} + 1} + \ldots + X_{\tau_1 + \ldots + \tau_k}$

(where ${S_1 = X_1 + \ldots + X_{\tau_1} \equiv X_1 + \ldots + X_\tau}$). Exactly the same argument as above can be used to show that ${S_1,S_2,\ldots}$ are i.i.d.. Now, noting that $\displaystyle S_1 + \ldots + S_k = X_1 + X_2 + \ldots + X_{\tau_1 + \ldots + \tau_k},$

we can write $\displaystyle \frac{S_1 + \ldots + S_k}{k} = \frac{X_1 + \ldots + X_{\tau_1 + \ldots + \tau_k}}{k} = \frac{X_1 + \ldots + X_{\tau_1 + \ldots + \tau_k}}{\tau_1 + \ldots + \tau_k} \cdot \frac{\tau_1 + \ldots + \tau_k}{k}.$

Let’s see what happens as ${k \rightarrow \infty}$. First of all, if we let ${Z_n := n^{-1}(X_1 + \ldots + X_n)}$, then, by the SLLN, ${Z_n \rightarrow \mathop{\mathbb E} X_1 =\mu}$ a.s. as ${n \rightarrow \infty}$. Since any subsequence of a convergent sequence converges to the same limit, we have $\displaystyle \lim_{k \rightarrow \infty} \frac{X_1 + \ldots + X_{\tau_1 + \ldots + \tau_k}}{\tau_1 + \ldots + \tau_k} = \lim_{k \rightarrow \infty} Z_{\tau_1 + \ldots + \tau_k} = \mu, \qquad \text{a.s.}$

Moreover, since ${\tau_1,\tau_2,\ldots}$ are i.i.d. with mean ${\mathop{\mathbb E} \tau}$, ${k^{-1}(\tau_1 + \ldots + \tau_k) \rightarrow \mathop{\mathbb E}\tau}$ almost surely as ${k \rightarrow \infty}$. Therefore, $\displaystyle \lim_{k \rightarrow \infty} \frac{S_1 + \ldots + S_k}{k} = \mu \cdot \mathop{\mathbb E}\tau \qquad \text{a.s.}$

Now we invoke the converse to the SLLN (more precisely, the contrapositive to the converse): Since ${S_1,S_2,\ldots}$ are i.i.d. and the limit of ${k^{-1}(S_1+\ldots+S_k)}$ exists a.s., then ${\mathop{\mathbb E} S_1}$ is finite and equal to this limit. Hence, $\displaystyle \mathop{\mathbb E} (X_1 + \ldots + X_\tau) = \mathop{\mathbb E} S_1 = \mu \cdot \mathop{\mathbb E} \tau.$

The theorem is proved.

### 4 Responses

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1. Vijay Subramanian said, on April 29, 2011 at 11:56 am

Seems like a renewal-reward theorem argument with the \tau sequence being the renewal epochs and the S sequence being the rewards that one gets.

• mraginsky said, on April 29, 2011 at 12:02 pm

That’s certainly a nice take on it.

2. Anand Sarwate said, on April 29, 2011 at 4:16 pm

Ah so cute!

3. Linkage « An Ergodic Walk said, on May 1, 2011 at 11:36 pm

[…] Blackwell’s proof of Wald’s Identity, as told by Max. […]