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Suppose X is a continuous random variable that can take any value between plus and minus infinity. Furthermore, suppose A is a random variable capturing those events where X is below 0, and B is a random variable capturing those events where X is above 0.

Is there a general relationship between variance(X), variance(A), and variance(B)?enter image description here

UPDATE: Siong and Canardini provide the same answer. Unfortunately, my simulation efforts do not agree, so I wonder where the mistake is:

In the following I draw 10 numbers (MATLAB, randn) and try to implement the given answers. What am I doing wrong?

enter image description here

user120911
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2 Answers2

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$$X=A+B$$

$$Var(X)=Var(A)+Var(B)+2Cov(A,B)$$

$$Cov(A,B)=\mathbb{E}[AB]-\mathbb{E}[A]\mathbb{E}[B]=-\mathbb{E}[A]\mathbb{E}[B]\geq 0$$

$$Var(X)=Var(A)+Var(B)-2\mathbb{E}[A]\mathbb{E}[B]$$

$$Var(A)+Var(B) \leq Var(X)$$

Siong Thye Goh
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  • Thank you, Siong. Unfortunately, while both you and Canardini provide the same answer, my simulations in MATLAB do not produce the right answer. See my update for details. Can you spot my mistake? – user120911 Jan 09 '17 at 22:52
  • In your computation of variance of $A$ and $B$, I think you did not consider those $0$ that are not printed out right? – Siong Thye Goh Jan 09 '17 at 22:57
  • I am not sure I follow, Siong. Can you please be a little more detailed? – user120911 Jan 09 '17 at 23:16
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    var([-0.13032, -0.79019, -0.93754]) gives us 0.1848, but var([-0.13032, -0.79019, -0.93754, zeros(1,7)]) gives us 0.1306. I think you just computed the variance of the non-zero parts while we included zero in the definition of $A$ and $B$. – Siong Thye Goh Jan 09 '17 at 23:25
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We have $$Var(X)=Var(A)+Var(B)+2Cov(A,B)$$

$$Cov(A,B)=E(AB)-E(A)E(B)$$

Given that $AB=X^21_{X \leq 0}1_{X >0}$, we have $AB=0$ a.s, $\{X \leq 0\}$ and $\{X >0\}$ being incompatible.

Thus, $$Var(X)=Var(A)+Var(B)-2E(A)E(B)$$

Canardini
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