Let X be a Bernoulli Random Variable whose parameter is a Uniform Random Variable Q which takes values in the domain [0 , 0.1].

We want to find var(X). My reasoning is the following:

Using the Law of Total Variance we have:

var(X) = E[var(X|Q)] + var(E[X|Q]) = E[Q * (1 - Q)] + var(Q) = E[Q] - E[Q^2] + var(Q) = E[Q] - (var(Q) + (E[Q])^2) + var(Q) = E[Q] - (E[Q])^2 = 0.05 - 0.05^2 = 0.0475.

Do you agree with this line of reasoning?

  • $\begingroup$ A Bernoulli random variable is either 0 with probability p or 1 with probability q=1-p. The possible value for the parameter belong to the interval [0,1] not [0, 0.1]. Its variance is p(1-p). $\endgroup$ – Michael R. Chernick Apr 19 '17 at 19:59
  • 2
    $\begingroup$ Please note that the interval [0, 0.1] is a subset of the interval [0, 1] and therefore it is a legitimate domain for the parameter. In the particular problem the parameter is modeled --as I stated-- as a Random Variable uniformly distributed over the interval [0, 0.1]. E.g. assume a biased coin. You know that its bias is between 0 --never Heads-- and 0.1 --10% of the times Heads. Then you model the parameter as a Random Variable uniformly distributed in the interval [0, 0.1]. In dealing with a parameter that is a Random Variable you have to use the Law of Total Variance. $\endgroup$ – rf7 Apr 19 '17 at 21:12
  • $\begingroup$ You are applying a distribution to an unknown fixed parameter. If you were doing Bayesian analysis this could make sense as a prior distribution. Then you would update when you observe a Bernoulli random outcome. $\endgroup$ – Michael R. Chernick Apr 19 '17 at 21:25
  • $\begingroup$ If you are just want to know the variance of a uniform random variable on [0,0,1] there is no need to bring the Bernoulli variable into the picture. You can just integrate 10 (the density) time (x-0.05)$^2 dx to get the variance. $\endgroup$ – Michael R. Chernick Apr 19 '17 at 21:31
  • $\begingroup$ Sorry, this was a typo. I want to find var(X) not var(Q) $\endgroup$ – rf7 Apr 19 '17 at 21:53

You are using the correct approach with the total variance law.

$$\mathbb{V}ar(X)= \mathbb{V}ar(\mathbb{E}(X|Q)) + \mathbb{E}(\mathbb{V}ar(X|Q) ).$$

Here you will need the expectations and variances of the Bernoulli and the continuous uniform. These are:

$$\mathbb{E}(X|Q)=Q,$$ $$\mathbb{V}ar(X|Q)=Q(1-Q),$$

$$\mathbb{E}(Q)=\frac{a+b}{2},$$ $$\mathbb{V}ar(Q)=\frac{(a-b)^2}{12},$$

where here $a=0.1$ and $b=0.0$, the limits of the continuous uniform. The only other thing I used to solve this was the variance relation

$$ \mathbb{V}ar(Y) = \mathbb{E}(Y^2)-(\mathbb{E}(Y))^2.$$

This is rearranged to get a relationship between the mean and the variance to get the second moment if you know the mean and the variance. So from the total variance law

$$\mathbb{V}ar(X)= \mathbb{V}ar(\mathbb{E}(X|Q)) + \mathbb{E}(\mathbb{V}ar(X|Q) ) =\mathbb{V}ar(Q) + \mathbb{E}(Q(1-Q)) =\mathbb{V}ar(Q) + \mathbb{E}(Q)-\mathbb{E}(Q^2) =\frac{0.1^2}{12} + 0.05 - (\frac{.01^2}{12} + 0.05^2) =0.0008333333 + 0.05 - (0.000833333 + 0.0025) =0.05 - 0.0025 =0.0475 $$ which matches what you have in the posting.


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.