First a warning: this is not the most interesting question but I want to update my understanding of independence now that I'm taking 1st year statistics

  • I often heard in my 1st year probability class that "independent RVs don't give any information about each other"

  • Rolling a fair dice $99$ times doesn't give info on the hundredth roll

  • However, let's say $X_1, ..., X_{100}$ are iid discrete uniform from $1$ to $\theta$ which is unknown

  • Then by observing $X_1,...,X_{99}$ you can make inferences about the hundredth observation and make a prediction interval

  • Is it safe to say that: given you know all the parameters or complete pmf/pdf of $F$, independent RVs don't give any information about each other. However, given that there are unknown parameters, independent RVs do give you information about each other. Does that make sense to even say?

I know this is more of an English question versus formal definitions (splitting pmfs/pdfs) but I'd like to try and be precise about this. Thanks for your help and patience.

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    $\begingroup$ This is basically the question of whether you are taking a frequentist or Bayesian approach to modeling your dice rolls. Do you think of the unknown parameter as "determined but not known to you", or "random"? $\endgroup$ Commented Jun 13, 2019 at 17:09
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    $\begingroup$ Yes, you are correct. In the case when $\{X_1, ..., X_{100}\}$ all depend on a common random variable $\theta$ then you can say they are conditionally i.i.d. given $\theta$, but they are not i.i.d. because they all depend on the random $\theta$. [edit: In view of the above comment, you can say I am taking a "Bayesian" approach to interpreting $\theta$ itself as a random variable.] $\endgroup$
    – Michael
    Commented Jun 13, 2019 at 17:10
  • $\begingroup$ @NateEldredge Thanks for the info... I'm aware of the Frequentist vs Bayesian debate but I'm not at a level to fully appreciate it. $\endgroup$ Commented Jun 13, 2019 at 17:44

1 Answer 1


Note the Nate Eldredge comment on two different views of the unknown $\theta$.

1) If $\theta$ is a random variable then you can say $\{X_1, ..., X_{100}\}$ are conditionally i.i.d. given $\theta$, but they are not i.i.d. because they all depend on the common random variable $\theta$. This is the "Bayesian" view of $\theta$.

2) If $\theta$ is a fixed (but possibly unknown) constant then indeed you can say $\{X_1, ..., X_{100}\}$ are i.i.d. because \begin{align} &P[X_1\leq x_1, ..., X_{100}\leq x_{100}] \\ &= \prod_{i=1}^{100}P[X_i\leq x_i] \quad \forall (x_1, ..., x_{100}) \in \mathbb{R}^{100} \quad (Eq. 1) \end{align} It can be shown that (Eq. 1) implies $$ E[h(X_{100})|X_1,...,X_{99}]=E[h(X_{100})] \quad (Eq. 2)$$ for all (measurable) functions $h$. The equations (Eq. 1) and (Eq. 2) hold regardless of whether or not the value $\theta$ is known. In particular, the expressions in (Eq. 1)-(Eq. 2) may depend on $\theta$, but the expressions exist (and the left-hand-sides are equal to the right-hand-sides) regardless of whether or not $\theta$ is known/unknown to an observer.

You can indeed interpret the Equations (1)-(2) to mean "the variables provide no information about each other": Knowing the outcomes of $X_1, ..., X_{99}$ does not change probabilities or expectations that involve only $X_{100}$. Of course, those probabilities/expectations are themselves unknown if $\theta$ is not known. So at a "higher level" you can indeed say that $X_1, ..., X_{99}$ gives "information" about the unknown $\theta$ and hence "information" about $X_{100}$. For example, if we observe $X_{99}=207$ then we know it is possible for $X_{100}>200$. However, this does not change $P[X_{100}>200]$ because that probability itself depends on $\theta$. It is difficult to quantify what "information" means without taking the "Bayesian" approach of treating the unknown $\theta$ itself as a random variable.

On the other hand, there are some interesting things that can be said about $\theta$ when we just treat it as a constant (not a random variable), such as mean-square-error of approximating $\theta/2$: If $\{X_i\}_{i=1}^{\infty}$ are i.i.d. uniform over $[0,\theta]$ then $$ E\left[\left(\frac{\theta}{2} - \frac{1}{n}\sum_{i=1}^n X_i\right)^2\right] = \frac{Var(X_1)}{n} = \frac{\theta^2}{12 n}$$ Of course this bound itself depends on $\theta$, but if we somehow know that $\theta \leq 100$ then we can say the mean-square-error is no more than $100^2/(12n)$.

Here is an estimator of $\theta$ with an improved mean-square-error: Define \begin{align} \hat{\theta}_n &= \frac{2}{n}\sum_{i=1}^n X_i\\ \tilde{\theta}_n &= \max\left\{\hat{\theta}_n, X_1, X_2, ..., X_n\right\} \end{align} It can be shown that (surely): $$ (\tilde{\theta}_n-\theta)^2 \leq (\hat{\theta}_n -\theta)^2$$ and so $$ E\left[\left(\tilde{\theta}_n-\theta\right)^2\right] \leq E\left[\left(\hat{\theta}_n-\theta\right)^2\right] = \frac{\theta^2}{3n}$$ Some useful notes on other improvements are here:


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    $\begingroup$ Thanks for your help Michael and clearly pointing out the distinction when the unknown parameter is a RV or a fixed constant. $\endgroup$ Commented Jun 13, 2019 at 18:17
  • $\begingroup$ I also appreciate your insight on the language part of "information." In regards to your final comment, when $\theta$ is a constant for iid uniform discrete... it seems like you're saying that the bound for MSE is pretty good and so if $n$ is large then you can have a high confidence level on approximating the expected value... and so a big $n$ means that the iid RVs give you a lot of "information" about future values... is that the right way to think about it? $\endgroup$ Commented Jun 13, 2019 at 18:21
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    $\begingroup$ Yes, "confidence" is a good word. I added a paragraph in my answer on a simple improved estimator of $\theta$. $\endgroup$
    – Michael
    Commented Jun 14, 2019 at 16:52
  • $\begingroup$ Thanks for your help Michael, I appreciate it! Yes I remember the discrete uniform was used as an example in my course of a distribution where the typical sample mean is not the best estimator which I thought was interesting $\endgroup$ Commented Jun 14, 2019 at 17:07
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    $\begingroup$ My calculations of $\theta^2/(3n)$ are for the continuous uniform case, based on $E[X_1]=\theta/2$, $Var(X_1) = \theta^2/12$. Similar holds for the discrete uniform but the variance calculation is a bit different. $\endgroup$
    – Michael
    Commented Jun 14, 2019 at 17:08

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