2
$\begingroup$

Let $N$ random variable taking positive integer values, with mean $a$ and variance $r$. Let $X_i$ i.i.d. random variables with mean $b$ and variance $s$. Let $Y_i$ i.i.d random variables with mean $c$ and variance $t$. Let $N$, $X_i$, $Y_i$ independent. Let $A = \sum_{i=1}^NX_i$ Let $B = \sum_{i=1}^NY_i$

What is $\mathbb{V}ar(A + B)$?

I reason as follows:

$\mathbb{V}ar(A + B) = \mathbb{V}ar(A) + \mathbb{V}ar(B)$

It can be shown that the variance of a random sum of independent random variables $X_i$ ($i = 1,...,N$) is:

$$ \mathbb{V}ar(A) = E[N]*\mathbb{V}ar(X) + (E[X])^2 * \mathbb{V}ar(N) = a * s + b^2 * r$$ similarly $$ \mathbb{V}ar(B) = \mathbb{E}[N] * \mathbb{V}ar(Y) + (\mathbb{E}[Y])^2 * \mathbb{V}ar(N) = a * t + c^2 * r$$

Therefore, $$\mathbb{V}ar(A + B) = \mathbb{V}ar(A) + \mathbb{V}ar(B) = a * s + a * t + b^2 * r + c^2 * r = a * (s + t) + r * (b^2 + c^2)$$ However, this is not the right answer. Could you help me understand what I am doing wrong? Your advice will be appreciated.

$\endgroup$
2
  • 1
    $\begingroup$ The variance depends on the covariance between A and B. $\endgroup$
    – SmallChess
    Apr 19, 2017 at 5:45
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Variance. Skip to "Weighted sum of variables". $\endgroup$
    – SmallChess
    Apr 19, 2017 at 5:45

1 Answer 1

1
$\begingroup$

Let's see: by the Law of Total Variance

$$\text {Var}(A+B) = \text {Var}[E(A+B\mid N)]+ E[\text {Var}(A+B\mid N)]$$

We have

$$E(A+B\mid N) = N(b+c) \implies \text {Var}[E(A+B\mid N)] = (b+c)^2\cdot \text {Var}(N)$$

and

$$\text {Var}(A+B\mid N) = N(s+t) \implies E[\text {Var}(A+B\mid N)] = E(N)\cdot (s+t)$$

So

$$\text {Var}(A+B) = (b+c)^2\cdot r+ a\cdot (s+t)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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