June 17th, 2021, 6:42 pm
Friends, I wanted to share the analytics for multidimensional monte carlo simulations of the kind we have in equity basket pricing where each of the N SDEs in the basket of N assets is largely single-dimensional but their driving brownian motions are correlated.
I suppose that we have already done cholesky decomposition or SVD decomposition of the correlation matrix and we have written each of the N SDEs with their volatility coefficients explicitly stated after cholesky or SVD. I will just write one of the N SDEs in the basket as
[$]dX_1 = \mu_1 \, {X_1}^{\beta_1} \, dt + \sum_{n=1}^{N} \, \sigma_{1n} \, {X_1}^{\gamma_1} \, dZ_n[$]
In particular we want to notice that [$]\sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, = \, {\sigma_1}^2 [$]
Please note that after cholesky or SVD, all of the above brownian motions [$]Z_n[$] are orthogonal.
Since I am playing with only one SDE, I will drop the subscript "1" from the bottom of each variable since it is unnecessary in case we are dealing with one SDE only.
expanding the above SDE as we have previously done, we write the equation after applying repeated Ito as
[$]dX = \mu \, {X}^{\beta} \, \int_0^t ds + \, {X}^{\gamma} \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t dZ_n(s) [$]
[$]+(\mu \, \beta {X}^{\beta-1} ) \, (\mu \, {X}^{\beta}) \, \int_0^t \int_0^s dv \, ds[$]
[$]+(\mu \, \beta {X}^{\beta-1} ) \, ({X}^{\gamma} ) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_n(v) ds[$]
[$]+.5 (\mu \, \beta \, (\beta-1) \, {X}^{\beta-2} ) \, ({X}^{2 \gamma} ) \, \sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, \int_0^t \int_0^s dv \, ds[$]
[$]+(\gamma {X}^{\gamma-1} ) \, (\mu \, {X}^{\beta}) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]
[$]+(\gamma {X}^{\gamma-1} ) \, ({X}^{\gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, \sigma_{1m} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_m(v) \, dZ_n(s)[$]
[$]+.5 (\gamma (\gamma-1) \, {X}^{\gamma-2} ) \, ({X}^{2 \gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, {\sigma_{1m}}^2 \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]
since in quadratic variations, [$]\sum_{n=1}^{N} \, {\sigma_{1n}}^2 \, = \, {\sigma_1}^2 [$], we can slightly simplify the above equations by using this identity that changes double summations in quadratic variations to single summations and write the above equation again as
[$]dX = \mu \, {X}^{\beta} \, \int_0^t ds +\, {X}^{\gamma} \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t dZ_n(s) [$]
[$]+(\mu \, \beta {X}^{\beta-1} ) \, (\mu \, {X}^{\beta}) \, \int_0^t \int_0^s dv \, ds[$]
[$]+(\mu \, \beta {X}^{\beta-1} ) \, ({X}^{\gamma} ) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_n(v) ds[$]
[$]+.5 (\mu \, \beta \, (\beta-1) \, {X}^{\beta-2} ) \, ({X}^{2 \gamma} ) \, {\sigma_1}^2 \, \int_0^t \int_0^s dv \, ds[$]
[$]+(\gamma {X}^{\gamma-1} ) \, (\mu \, {X}^{\beta}) \, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]
[$]+(\gamma {X}^{\gamma-1} ) \, ({X}^{\gamma}) \, \sum_{n=1}^{N} \, \sum_{m=1}^{N} \, \sigma_{1m} \, \sigma_{1n} \, \int_0^t \int_0^s dZ_m(v) \, dZ_n(s)[$]
[$]+.5 (\gamma (\gamma-1) \, {X}^{\gamma-2} ) \, ({X}^{2 \gamma}) \, {\sigma_1}^2\, \sum_{n=1}^{N} \, \sigma_{1n} \, \int_0^t \int_0^s dv \, dZ_n(s)[$]
We can solve all of the above integrals analytically.
[$]\int_0^t \int_0^s dZ_n(v) ds= \frac{1}{\sqrt{3}} t Z_n(t)[$]
[$]\int_0^t \int_0^s dZ_n(v) dZ_n(s)= \frac{1}{2} ({Z_n(t)}^2-t)=H_2(Z_n(t))[$]
only difficult integral is
[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]
where we get this integral from Ito Isometry as
[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)[$]
[$]=\int_0^t Z_m(s) dZ_n(s)[$]
and its variance is given as
[$]=\int_0^t {Z_m(s)}^2 ds[$]
[$]=\int_0^t d[{Z_m(s)}^2 s]-\int_0^t s d[{Z_m(s)}^2][$]
[$]= t \, {Z_m(t)}^2 - \int_0^t 2 s \, Z_m(s) \, dZ_m(s)- \int_0^t s \, ds [$]
[$]= t \, H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t^2/2[$]
its representation will be given as
[$] \sqrt{(H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t/2)} \, \sqrt{t} \, N_n[$]
where [$]N_n[$] is standard normal associated with brownian motion [$]Z_n[$]
writing [$]\sqrt{t} \, N_n = Z_n(t)[$] in the above equation we get
[$] \sqrt{(H_2(Z_m(t)) (1-\frac{\sqrt{2}}{2}) + t/2)} \, Z_n(t)[$]
[$]= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]
So we can write the integral as
[$]\int_0^t \int_0^s dZ_m(v) dZ_n(s)= Z_n(t) \sqrt{[{Z_m(t)}^2-t] (1-\frac{\sqrt{2}}{2}) +\frac{t}{2} }[$]
We can now solve all the integrals and easily simulate a basket option to 2nd expansion order of Ito-Taylor expansion.
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