Exchanging expectation and limits

I have a stochastic process, ${b_t} \, (t=0, 1, 2, \ldots)$, which follows a random walk. Specifically, ${b_0} = 0$ and for $t$ greater than zero, $\displaystyle {b_t} = \sum\limits_{i = 1}^t {{\varepsilon _i}} $.

The ${\varepsilon _i}$ are i.i.d. random variables with zero expected value. Other than the existence of the expectation, I would rather not have to assume any stronger regularity on the ${\varepsilon _i}$.

Let $0 < \beta < 1$ (beta is a discount factor).

My question is: Is it the following true? $${E_{t = 0}}\left( {\mathop {\lim }\limits_{t \to \infty } {\beta ^t}{b_t}} \right) = 0$$

By ${E_{t = 0}}$ I mean the expected value at time zero. It seems to me that the proof depends on whether I can exchange the limit and the expectation, since $\forall t, {E_{t = 0}}\left( {{b_t}} \right) = 0.$ Please, provide the details of the proposed proof, since I am not a mathematician (just an economist). Thank you.

Thank you!

By working in the first two answers received I realized that the nature of my problem (stating a meaningful transversality condition for an optimization problem, and proving that the proposed solution meets it) required me to prove that ${\beta ^t}{b_t}\mathop \to \limits^{a.s.} 0$ and not only that ${E_{t = 0}}\left( {\mathop {\lim }\limits_{t \to \infty } {\beta ^t}{b_t}} \right) = 0$.

I think that the third answer, based on the Strong Law of Large Numbers, kills both birds with the same stone, and a clear and elegant stone to boot. Thanks to all.

  • $\begingroup$ $var (\beta b_t) = \sigma_{\epsilon}^2\beta ^{2t}t$, hence $\beta b_t \overset{p}{\to} 0$. $\endgroup$
    – Math-fun
    Jun 30 '15 at 21:43

By Strong Law of Large Numbers, without any assumptions on the variance of $\epsilon_i$, $$ \frac{1}{n}\sum^n_{i=1}\epsilon_i \rightarrow \mathbb E[\epsilon_i]=0 \quad \text{almost surely} $$ So since $\lim_{n \rightarrow \infty} n\beta^n = 0$ and $n^{-1}b_n \rightarrow 0$ almost surely, we have that $$ \beta^n b_n\rightarrow 0 \quad \text{almost surely} $$ Now $$ \mathbb E \left[\lim_{n\rightarrow \infty} \beta^n b_n\right] = \mathbb E [0]=0 $$ since $\lim_{n\rightarrow \infty} \beta^n b_n=0$ on the whole probability space except on a set of measure $0$.

So this is done without any interchanging of limits and expectations.


Assuming that $Var(\epsilon_1)=\sigma_{\epsilon}^2$ we have

$$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$

Hence, $\beta^tb_t$ is $L^2$-bounded and uniformly integrable so that you can pass the limit inside the expectation...


In the first step, you need to show that $\beta^tb_t$ converges a.s. (to $0$). This can be done in various ways, e.g. using Kolmogorov's two-series theorem or the SLLN.


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