I've got the following random variable for which I must find the expected value and variance:
$X_h =\sum_{i=1}^{15} X_i$
Where $X_i$ is a random variable of the set $s = \{0, 1, 3\}$, corresponding to the probability density function $f(x) = \left\{ \begin{array}{ll} 0.6 & \mbox{if } x = 3 \\ 0.2 & \mbox{otherwise} \end{array} \right.$.

I got the expected value by simply stating that:

$E(X_h) = E(\sum_{i=1}^{15} X_i) = \sum_{i=1}^{15} E(X_i) = \sum_{i=1}^{15} (\sum_s x\cdot f_x(x)) =\sum_{i=1}^{15} (0\cdot 0.2 + 1 \cdot 0.2 + 3 \cdot 0.6) = 30$

I did the following for the variance (which appears to be incorrect):
$Var(X_h) = E(X_h^2)-E(X_h)^2 = E((\sum_{i=1}^{15} X_i)\cdot(\sum_{i=1}^{15} X_i)) - E(X_h)^2$

$=E(\sum_{i=1}^{15}\sum_{j=1}^{15} X_i\cdot X_j) - E(X_h)^2$

$=\sum_{i=1}^{15}\sum_{j=1}^{15} E(X_i\cdot X_j) - E(X_h)^2$

$= \sum_{i=1}^{15}\sum_{j=1}^{15} (\sum_s X_i\cdot X_j \cdot f_x(X_i)) - E(X_h)^2$

$= \sum_{i=1}^{15}\sum_{j=1}^{15} (\sum_s X_i^2 \cdot f_x(X_i)) - E(X_h)^2$

$= \sum_{i=1}^{15}\sum_{j=1}^{15} (3^2 \cdot 0.6+1^2\cdot 0.2 + 0^2\cdot 0.2) - E(X_h)^2$

$= 15^2(3^2 \cdot 0.6+1^2\cdot 0.2 + 0^2\cdot 0.2) - 30^2$

$= 360$

Now my solution paper claims the following:
$Var(X_h) = \sum_{i=1}^{15} Var(X_i) = 15 \cdot 1.6 = 24$

I've been using this answer to compute the variance here but I think there may be a flaw somewhere. Is correct that even if the variables are independent, that calculating variance formally should yield the same result? What did I do wrong?

  • $\begingroup$ I don't understand your definition of distribution of the $X_i$. $\endgroup$
    – Glen_b
    Jun 3, 2015 at 9:14
  • $\begingroup$ @Glen_b it's a random variable of the set s with a pdf given above. That's all. The index doesn't really mean anything. $\endgroup$ Jun 3, 2015 at 9:16
  • $\begingroup$ The term "Bernoulli" normally refers to a distribution over the values {0,1}, while your definition appears to be a discrete distribution over 3 values, but it's not clearly enough described to be certain (since there's a slight ambiguity in your description of the probability function). If you remove the term "Bernoulli" (or clarify enough to justify it) it may be clear enough. $\endgroup$
    – Glen_b
    Jun 3, 2015 at 9:39

1 Answer 1


The problem is that you should not write:

$$E(X_i.X_j) = \sum_s X_i.X_j.f_x(X_i)$$

Because the r.v $X_i.X_j$ follows its joint distribution, not the marginal distribution.

  • $\begingroup$ Then what should be written instead? $\sum_s X_i \cdot X_j f_x(X_i, X_j)$? Also how do I summate this? Could you provide an example? $\endgroup$ Jun 3, 2015 at 9:25
  • $\begingroup$ $\sum_s x_i.x_j.f_{X_i,X_j}(x_i,x_j)$ and you need to calculate the joint distribution. $\endgroup$
    – Metariat
    Jun 3, 2015 at 9:28
  • $\begingroup$ @Xuan_Quang_DO: since they're independent the joint probability distribution is: $f_{X_i}(x_i)\cdot f_{X_j}(x_j)$. Now for what variables does the sum go? Does it become a double sum like: $\sum_{x_i=\{0,1,3\}} \sum_{x_j=\{0,1,3\}} x_i x_j f_{X_i}(x_i) f_{X_j}(x_j)$ ? $\endgroup$ Jun 3, 2015 at 9:31
  • $\begingroup$ Yes, but I didn't see the independent hypothesis in your question. And in the independent case, the Var of the sum is simply the sum of Var! $\endgroup$
    – Metariat
    Jun 3, 2015 at 9:36
  • $\begingroup$ @Xuan_Quang_DO: Yes, that is what the original answer states as well. However, when I summate according to the above and subtract the expected value ($E(X_h^2)-E(X_h)^2$) I get 0. Why is that? $\endgroup$ Jun 3, 2015 at 9:40

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.