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.


1 Answer 1


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 $X$.

  • $\begingroup$ 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. $\endgroup$
    – mlofton
    Oct 8, 2021 at 2:00

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