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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.

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

1 Answer 1

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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

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

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