How Skewness Scales Across Independent Returns and Forward Periods

# Skewness Equivalent to Additivity of Variance


# Skewness Equivalent to Additivity of Variance












Let $S$ be a diffusion with independent increments and suppose we have options on $S$ with expiries $T_1$ and $T_2$ and ATM implied vols $\sigma_1$ and $\sigma_2$. Let $X_1$ and $X_2$ denote the log returns of $S$ over these two periods. Then, since the variance of independent random variables is additive, we can construct forward vols via $$\tilde{\sigma_2} = \sqrt{\frac{\sigma_2^2 T_2 - \sigma_1^2 T_1}{T_2-T_1}}$$ Now, denote $\kappa_1 = E[X_1^3]$ and $\kappa_2 = E[X_2^3]$. Assuming 0 mean, by independence we have $$E[\log(S_2/S_0)^3] = \kappa_1 + \kappa_2$$ Using this result, I want to derive a forward skew using some measurement of the skew. How might these quantities scale with time? In other words, is there an equivalent statement to variance scaling linearly with time? Purely off intuition, my guess is $\propto \sqrt{T^3}$.

Follow up question: What if we made these properties scale invariant and divided by the ATM vols? Is there an alternative relationship in this case?

Looking for some results on independent increment processes to disprove or justify my intuition.

## Answer by Kermittfrog (score 2, accepted)

https://quant.stackexchange.com/a/79559

Update: I think i misunderstood your question, so let me add another answer as well.

Answer A will derive a formula for forward skewness, answer B will show the general property of skewness as a function of the number of increments that we are adding.

### Answer A: Forward skewness

Assume two independent mean-zero increments $X_1$ and $X_2$ and let their sum be $Z=X_1+X_2$. We further assume knowledge of $\mathrm{Var}(X_1)$, $\mathrm{skew}(X_1)$ as well as $\mathrm{Var}(Z)$ and $\mathrm{skew}(Z)$ already exists.

We can then rewrite the skewness equation $$ \begin{align} \mathrm{skew}(Z)&\equiv \frac{\mathrm{E}(X_1^3+X_2^3)}{\mathrm{E}\left(X_1^2+X_2^2\right)^{1.5}}\\ &= \frac{\mathrm{skew}(X_1)\mathrm{Var}(X_1)^{1.5}+\mathrm{skew}(X_2)\mathrm{Var}(X_2)^{1.5}}{\left(\mathrm{Var}(X_1)+\mathrm{Var}(X_2)\right)^{1.5}}\\ \Rightarrow \mathrm{skew}(X_2)&=\frac{\mathrm{skew}(Z)\left(\mathrm{Var}(X_1)+\mathrm{Var}(X_2)\right)^{1.5}-\mathrm{skew}(X_1)\mathrm{Var}(X_1)^{1.5}}{\mathrm{Var}(X_2)^{1.5}}\\ &=\frac{E(Z^3)-E(X_1^3)}{(E(Z^2)-E(X_1)^2)^{1.5}} \end{align} $$

Put differently, as $E(Z^k)=E(X_1^k)+E(X_2^k)$, we can simply solve for both forward moments $E(X_2^k)$ ($k=2$,$k=3$) and calculate forward skewness rather simply. Same holds for kurtosis, of course.

### Answer B: Skewness of $n$ independent increments

With independent increments $X_i$ and given the standardised moments

$$ \begin{align} E(X_i)&\equiv k_1\\ E\left(\left(X_i-E(X_i)\right)^2\right)&\equiv k_2\\ E\left(\left(X_i-E(X_i)\right)^3\right)&\equiv k_3 \end{align} $$

we have for the random variable $Z=\sum_{i=1}^nX_i$ $$ \begin{align} E(Z)&= nk_1\\ E\left(\left(Z-E(Z)\right)^2\right)&= nk_2\\ E\left(\left(Z-E(Z)\right)^3\right)&= nk_3 \end{align} $$

i.e. the standardised moments scale linearly with the number of increments (i.e. with time). As skewness is defined as $\mathrm{skew}\equiv \frac{k_3}{k_2^{1.5}}$ we have

$$ \mathrm{skew}(Z)=\frac{E\left(\left(Z-E(Z)\right)^3\right)}{\left(E\left(\left(Z-E(Z)\right)^2\right)\right)^{1.5}}=\frac{n}{n^{1.5}}\frac{k_3}{k_2^{1.5}}=\frac{1}{\sqrt{n}}\mathrm{skew}(X) $$

I.e. theoretically, with independent increments, skewness will tend to zero roughly at $\frac{1}{\sqrt{N}}$.

## Answer by Arshdeep (score 1)

https://quant.stackexchange.com/a/79525

$Y=log(S2/S1)=X2-X1$.

$E(Y+X1)^3=skew(Y)+k1=k2$, so $skew(Y)=k2-k1.$

If we assume same skew for all $log (S_k/S_{k-1})$ then skew grows linearly with time, just as variance.

If you can describe what you mean by making properties scale invariant, I can help you with that as well.

Thanks!