Let $Z_1 , Z_2 , \ldots$ be iid with $E(|Z_i|) < \infty$. Let $\theta$ be independent of the $Z_i$ with $E(|\theta|) < \infty$. Let $Y_i = \theta + Z_i$. I am trying to show that the conditional expectations $E(\theta\mid Y_1 ,\ldots, Y_n)$ converge a.s. to $\theta$. I know that $E(\theta\mid Y_1 ,\ldots , Y_n) \rightarrow E(\theta \mid \sigma(Y_1, Y_2 , \ldots))$ a.s. as $n\rightarrow \infty$, so it will suffice to show that $E(\theta \mid \sigma(Y_1, Y_2 , \ldots)) = \theta$. (here $\sigma(Y_1 , Y_2 , \ldots)$ is the $\sigma$ algebra generated by the $Y_i$) However, I do not see a way to do this. Any suggestions?


By the Strong Law of Large Numbers, there's a set $\Omega$ of measure 1 on which $$ \lim_{n\to\infty}\frac{Z_1+\cdots+Z_n}{n} = \mathbb{E}(Z_1).$$

So, on this set, we have

$$ \theta+ \lim_{n\to\infty}\frac{Z_1+\cdots+Z_n}{n} = \theta + \mathbb{E}(Z_1).$$

That is,

$$ \lim_{n\to\infty}\frac{(\theta+Z_1)+\cdots+(\theta+Z_n)}{n} = \theta + \mathbb{E}(Z_1).$$


$$ \lim_{n\to\infty}\frac{Y_1+\cdots+Y_n}{n} = \theta + \mathbb{E}(Z_1).$$

Since $\theta$ is (a.s.) equal to a $\sigma(Y_1,Y_2,\cdots)$-measurable random variable,

$$\mathbb{E}(\theta|\sigma(Y_1,Y_2,\cdots)) = \theta \text{ a.s.}$$

so we're done.

  • 1
    $\begingroup$ Did you need to use the $\theta$ independence here? I am guessing you implicitly use it in the final one to get $Y_{i}\sim N(\theta,1)$ are iid conditionally on $\theta$ and then apply SLLN on them. $\endgroup$ – OOESCoupling yesterday
  • $\begingroup$ @OOESCoupling, no. The argument applies the SLLN once to $Z_1,Z_2,\cdots$. $\endgroup$ – Ben Derrett yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.