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I am trying to determine the variance of the payout of a straddle. For puts and calls individually:

Var[P] = E[P^2] - E[P]^2
Var[C] = E[C^2] - E[C]^2

where: $$ E[P] \text = e^{-r T}\int _0{}^k\frac{(k-S) }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS $$

$$ E[C] \text = e^{-r T}\int _k{}^{\infty}\frac{(S-k) }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS $$ $$ E[P^2] \text = e^{-2r T}\int _0{}^k\frac{(k-S)^2 }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS $$

$$ E[C^2] \text = e^{-2r T}\int _k{}^{\infty}\frac{(S-k)^2}{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS $$ Where: $$ a=T \left(r-\frac{\sigma ^2}{2}\right)+\log \left(S_0\right) $$ and $$ b=\sigma \sqrt{T} $$

These results become:

$$ E[P] = k \Phi \left(-d_2\right) e^{-r T}-S_0 \Phi \left(-d_1\right) $$ $$ E[C] = S_0 \Phi \left(d_1\right)-k \Phi \left(d_2\right) e^{-r T} $$ $$ E[P^2] = k^2 \Phi \left(-d_2\right) e^{-2 r T}-2 k S_0 \Phi \left(-d_1\right) e^{-r T}+S_0^2 \Phi \left(d_2-2 d_1\right) e^{\sigma ^2 T} $$ $$ E[C^2] = k^2 \Phi \left(d_2\right) e^{-2 r T}-2 k S_0 \Phi \left(d_1\right) e^{-r T}+S_0^2 \Phi \left(2 d_1-d_2\right) e^{\sigma ^2 T} $$

Note that $$ d_1=\frac{-\log (k)+T \left(r-\frac{\sigma ^2}{2}\right)+\log \left(S_0\right)+\sigma ^2 T}{\sigma \sqrt{T}} $$ and $$ d_2=d_1-\sigma \sqrt{T} $$

where:

S0 = initial price
S = price
k = strike
r = interest rate
T = time to expiration
sigma = implied volatility
Phi = standard normal variable

These results are also shown in this paper: https://arxiv.org/pdf/1204.3452.pdf

My question: how to combine Var[P] and Var[C] to get Var[straddle]?

Since C and P are negatively correlated(right?), can we use this? $$ Var[straddle] = Var[P] + Var[C] - 2\sqrt{Var[P] Var[C] } $$

This seems alright, I'm not sure. If we instead start with the payoff for the straddle as a piecewise function: $$ payoff_{straddle} = \begin{array}{cc} \{ & \begin{array}{cc} k-S & 0<S<k \\ S-k & S\geq k \\ \end{array} \\ \end{array} $$

then: $$ E[straddle] \text = e^{-r T}\int _0{}^{\infty}\frac{payoff_{straddle} }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS $$

But, after separating out the different pieces in the piecewise function, this becomes $$ E[straddle] \text = e^{-r T}(\int _0{}^k\frac{(k-S) }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS +\int _k{}^{\infty}\frac{(S-k) }{\sqrt{2 \pi } b S} e^{-\frac{(\log (S)-a)^2}{2 b^2}}dS) $$ which is simply $$ E[straddle] = E[C] + E[P] $$ and, similarly $$ E[straddle^2] =E[C^2] + E[P^2] $$ Var[straddle] would then be: $$ Var[straddle] = E[straddle^2] - E[straddle]^2 = E[C^2] + E[P^2] - E[P]^2 - E[C]^2 - 2E[C]E[P] $$ $$ Var[straddle] = Var[C] + Var[P] - 2 E[C]E[P] $$

The second approach seems more robust to me, since we are beginning with payoff for the position and building from there. Perhaps I am missing something though.Thoughts on correct method?

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Assuming that \begin{align*} C &= e^{-rT}(S-K)^+,\\ P &= e^{-rT}(K-S)^+, \end{align*} and \begin{align*} straddle = C+ P, \end{align*} then the formula \begin{align*} Var[straddle] = Var[C] + Var[P] - 2 E[C]E[P] \end{align*} is correct. However, the formula \begin{align*} Var[straddle] = Var[P] + Var[C] - 2\sqrt{Var[P] Var[C] } \end{align*} is wrong, as $C$ and $P$ are not perfectly negatively corrected. Specifically, since $CP=0$, \begin{align*} Covar(C, P) &= -E(C)E(P)\\ &\neq -\sqrt{Var[P] Var[C] }. \end{align*}

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  • $\begingroup$ Thank you. I knew that I was making a mistake somewhere, really appreciate you giving it a look for me. $\endgroup$ – pyrex Sep 16 '16 at 4:41

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