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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

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    The variance takes as an argument a random variable, not an event . Did you mean to define $A=X1_{X \leq 0}$ and $B=X1_{X > 0}$ ? – 2017-01-09
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    Yes, that is what I meant. Thank you, Canardini, for clarifying this. – 2017-01-09
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    You must fill the empty cells of Columns A and B with zeros – 2017-01-10

2 Answers 2

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$$X=A+B$$

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

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

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

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

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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? – 2017-01-09
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    In your computation of variance of $A$ and $B$, I think you did not consider those $0$ that are not printed out right? – 2017-01-09
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    I am not sure I follow, Siong. Can you please be a little more detailed? – 2017-01-09
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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 $B$. – 2017-01-09
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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, $$Var(X)=Var(A)+Var(B)-2E(A)E(B)$$