# Intuition: Variance of the sum of R.V's and correlation

I have many questions that seems basic to me but I just cannot wrap my head around it.

Say we simulate 100 R.Vs that comes from a symmetric distribution with mean 0 $$(X)$$. Say we build another random variable with 1-1 correspondence just by setting a minus in front of our values $$(-X)$$. Clearly they have same distribution and the correlation will be either 1 or -1.

Say we have a exponential function (f.x, stock price function in Black Scholes model), $$F(X)$$. First question will be, what is $$Var(F(X) + F(-X))$$? My intuition says, not 0 since we will still have values that varies from the mean? But.. $$Var(F(X)+F(-X))=2Var(F(X))+2Cov(F(X),F(-X)) = 2(Var(F(X))+\rho(F(X)+F(-X))*Var(F(X)))$$

Now if $$\rho(F(X)+F(-X)) =-1$$ the variance will be 0.

Will $$F(X)$$ and $$F(-X)$$ have same distributions? Because my lecture notes says yes, but I cannot wrap my head around why. If they have the same distributions, aren't we allowed to just say,

$$Var(F(X)+F(-X))=Var(2F(X))$$ ?

These seems like basic questions, but my mind must be rusty.

I guess there are several questions in your posts. So I'll make several small answers.

## You can't say that $$Var(F(X)+F(−X)) = Var(2F(X))$$.

You must be carefull not two mix distribution and random variables:
$$F(X)$$ and $$F(-X)$$ may have the same distribution but they are still two distinct random variables: they won't take the same values at the same times, they will just take these values with same probability. So it doesn't mean that $$F(X) + F(-X) = 2F(X)$$.

## $$F(X)$$ and $$F(-X)$$ do have the same distribution.

If two random variables $$X$$ and $$Y$$ have the same distribution, then, for any function $$f$$, $$f(X)$$ and $$f(Y)$$ will also have the same distribution. So in your case, $$F(X)$$ and $$F(-X)$$ do have the same distribution since $$X$$ is symmetrical around $$0$$.

## $$var(F(X) + F(-X))$$ is different from 0.

Your computation of the variance starts well : $$Var(F(X)+F(−X))=2Var(F(X))+2Cov(F(X),F(−X))$$ but then you cannot says that $$\rho(F(X), F(-X)) = -1$$.

Correlation isn't preserved by transformations of random variables : if $$\rho(X, Y) = -1$$ then you don't have that $$\rho(f(X), f(Y)) = -1$$.

Here, I guess you could say that $$cov(F(X), F(-X)) = E(F(X)F(-X)) - E(F(X))E(F(-X))$$
and thus as $$F(X)F(-X) = e^X e^{-X} = 1$$ and F(X) and F(-X) have same expectation (because they have same distibution) : $$cov(F(X), F(-X)) = 1 - E(F(X))^2 .$$ But I don't think you can give a precise value for this covariance without additionnal information on the distribution of $$X$$.

• Hi: Pohoua, your answer was great. OBIEK: I think what wasn't ( or isn't ) clear in your question was if you A) generate 100 X values and then multiplied them by negative one to create the 100 other X values or B) generate 100 X values and then generate another 100 X values and multiple the latter 100 by negative one. I think it's the second now but only after Pohoua pointed that out. Oct 8, 2021 at 2:00