Discretizing a Stochastic Volatility SDE
$begingroup$
How does the discrete time stochastic volatility model arise from the continuous time one?
I have the following continuous time stochastic volatility model. $S_t$ is the price, and $v_t$ is a variance process.
$$
dS_t = mu S_tdt + sqrt{v_t}S_t dB_{1t} \
dv_t = (theta - alpha log v_t)v_tdt + sigma v_t dB_{2t} .
$$
I'm more familiar with the discrete time version:
$$
y_t = exp(h_t/2)epsilon_t \
h_{t+1} = mu + phi(h_t - mu) + sigma_t eta_t \
h_1 sim Nleft(mu, frac{sigma^2}{1-phi^2}right).
$$
${y_t}$ are the log returns, and ${h_t}$ are the "log-volatilites." Keep in mind there might be some confusion about parameters; for example the $mu$s in each of these models are different.
How do I verify that the first discretizes into the second?
Here's my work so far. First I define $Y_t = log S_t$ and $h_t = log v_t$. Then I use Ito's lemma to get
begin{align*}
dY_t &= left(mu - frac{exp h_t}{2}right)dt + exp[h_t/2] dB_{1t}\
dh_t &= left(theta - alphalog v_t - sigma^2/2right)dt + sigma dB_{2,t}\
&= alphaleft(tilde{mu} - h_t right)dt + sigma dB_{2t}.
end{align*}
I got the state/log-vol process piece. I use the Euler method to discretize, setting $Delta t = 1$, to get
begin{align*}
h_{t+1} &= alpha tilde{mu} + h_t(1-alpha) + sigma eta_t \
&= tilde{mu}(1 - phi) + phi h_t + sigma eta_t \
&= tilde{mu} + phi(h_t - tilde{mu}) + sigma eta_t.
end{align*}
The observation equation is a little bit more difficult, however:
begin{align*}
y_{t+1} = Y_{t+1} - Y_t &= (mu - frac{v_t}{2}) + sqrt{v_t}epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2} right) + exp[ log sqrt{v_t}] epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2}right) + expleft[ frac{h_t}{2}right] epsilon_{t+1}.
end{align*}
Why is the mean return not $0$?
stochastic-processes stochastic-calculus stochastic-integrals sde
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
$endgroup$
add a comment |
$begingroup$
How does the discrete time stochastic volatility model arise from the continuous time one?
I have the following continuous time stochastic volatility model. $S_t$ is the price, and $v_t$ is a variance process.
$$
dS_t = mu S_tdt + sqrt{v_t}S_t dB_{1t} \
dv_t = (theta - alpha log v_t)v_tdt + sigma v_t dB_{2t} .
$$
I'm more familiar with the discrete time version:
$$
y_t = exp(h_t/2)epsilon_t \
h_{t+1} = mu + phi(h_t - mu) + sigma_t eta_t \
h_1 sim Nleft(mu, frac{sigma^2}{1-phi^2}right).
$$
${y_t}$ are the log returns, and ${h_t}$ are the "log-volatilites." Keep in mind there might be some confusion about parameters; for example the $mu$s in each of these models are different.
How do I verify that the first discretizes into the second?
Here's my work so far. First I define $Y_t = log S_t$ and $h_t = log v_t$. Then I use Ito's lemma to get
begin{align*}
dY_t &= left(mu - frac{exp h_t}{2}right)dt + exp[h_t/2] dB_{1t}\
dh_t &= left(theta - alphalog v_t - sigma^2/2right)dt + sigma dB_{2,t}\
&= alphaleft(tilde{mu} - h_t right)dt + sigma dB_{2t}.
end{align*}
I got the state/log-vol process piece. I use the Euler method to discretize, setting $Delta t = 1$, to get
begin{align*}
h_{t+1} &= alpha tilde{mu} + h_t(1-alpha) + sigma eta_t \
&= tilde{mu}(1 - phi) + phi h_t + sigma eta_t \
&= tilde{mu} + phi(h_t - tilde{mu}) + sigma eta_t.
end{align*}
The observation equation is a little bit more difficult, however:
begin{align*}
y_{t+1} = Y_{t+1} - Y_t &= (mu - frac{v_t}{2}) + sqrt{v_t}epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2} right) + exp[ log sqrt{v_t}] epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2}right) + expleft[ frac{h_t}{2}right] epsilon_{t+1}.
end{align*}
Why is the mean return not $0$?
stochastic-processes stochastic-calculus stochastic-integrals sde
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
$endgroup$
1
$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04
add a comment |
$begingroup$
How does the discrete time stochastic volatility model arise from the continuous time one?
I have the following continuous time stochastic volatility model. $S_t$ is the price, and $v_t$ is a variance process.
$$
dS_t = mu S_tdt + sqrt{v_t}S_t dB_{1t} \
dv_t = (theta - alpha log v_t)v_tdt + sigma v_t dB_{2t} .
$$
I'm more familiar with the discrete time version:
$$
y_t = exp(h_t/2)epsilon_t \
h_{t+1} = mu + phi(h_t - mu) + sigma_t eta_t \
h_1 sim Nleft(mu, frac{sigma^2}{1-phi^2}right).
$$
${y_t}$ are the log returns, and ${h_t}$ are the "log-volatilites." Keep in mind there might be some confusion about parameters; for example the $mu$s in each of these models are different.
How do I verify that the first discretizes into the second?
Here's my work so far. First I define $Y_t = log S_t$ and $h_t = log v_t$. Then I use Ito's lemma to get
begin{align*}
dY_t &= left(mu - frac{exp h_t}{2}right)dt + exp[h_t/2] dB_{1t}\
dh_t &= left(theta - alphalog v_t - sigma^2/2right)dt + sigma dB_{2,t}\
&= alphaleft(tilde{mu} - h_t right)dt + sigma dB_{2t}.
end{align*}
I got the state/log-vol process piece. I use the Euler method to discretize, setting $Delta t = 1$, to get
begin{align*}
h_{t+1} &= alpha tilde{mu} + h_t(1-alpha) + sigma eta_t \
&= tilde{mu}(1 - phi) + phi h_t + sigma eta_t \
&= tilde{mu} + phi(h_t - tilde{mu}) + sigma eta_t.
end{align*}
The observation equation is a little bit more difficult, however:
begin{align*}
y_{t+1} = Y_{t+1} - Y_t &= (mu - frac{v_t}{2}) + sqrt{v_t}epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2} right) + exp[ log sqrt{v_t}] epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2}right) + expleft[ frac{h_t}{2}right] epsilon_{t+1}.
end{align*}
Why is the mean return not $0$?
stochastic-processes stochastic-calculus stochastic-integrals sde
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
$endgroup$
How does the discrete time stochastic volatility model arise from the continuous time one?
I have the following continuous time stochastic volatility model. $S_t$ is the price, and $v_t$ is a variance process.
$$
dS_t = mu S_tdt + sqrt{v_t}S_t dB_{1t} \
dv_t = (theta - alpha log v_t)v_tdt + sigma v_t dB_{2t} .
$$
I'm more familiar with the discrete time version:
$$
y_t = exp(h_t/2)epsilon_t \
h_{t+1} = mu + phi(h_t - mu) + sigma_t eta_t \
h_1 sim Nleft(mu, frac{sigma^2}{1-phi^2}right).
$$
${y_t}$ are the log returns, and ${h_t}$ are the "log-volatilites." Keep in mind there might be some confusion about parameters; for example the $mu$s in each of these models are different.
How do I verify that the first discretizes into the second?
Here's my work so far. First I define $Y_t = log S_t$ and $h_t = log v_t$. Then I use Ito's lemma to get
begin{align*}
dY_t &= left(mu - frac{exp h_t}{2}right)dt + exp[h_t/2] dB_{1t}\
dh_t &= left(theta - alphalog v_t - sigma^2/2right)dt + sigma dB_{2,t}\
&= alphaleft(tilde{mu} - h_t right)dt + sigma dB_{2t}.
end{align*}
I got the state/log-vol process piece. I use the Euler method to discretize, setting $Delta t = 1$, to get
begin{align*}
h_{t+1} &= alpha tilde{mu} + h_t(1-alpha) + sigma eta_t \
&= tilde{mu}(1 - phi) + phi h_t + sigma eta_t \
&= tilde{mu} + phi(h_t - tilde{mu}) + sigma eta_t.
end{align*}
The observation equation is a little bit more difficult, however:
begin{align*}
y_{t+1} = Y_{t+1} - Y_t &= (mu - frac{v_t}{2}) + sqrt{v_t}epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2} right) + exp[ log sqrt{v_t}] epsilon_{t+1} \
&= left(mu - frac{exp h_t}{2}right) + expleft[ frac{h_t}{2}right] epsilon_{t+1}.
end{align*}
Why is the mean return not $0$?
stochastic-processes stochastic-calculus stochastic-integrals sde
stochastic-processes stochastic-calculus stochastic-integrals sde
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
edited Jan 12 at 5:20
Taylor
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
asked Jul 9 '18 at 18:47
TaylorTaylor
287113
287113
1
$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04
add a comment |
1
$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04
1
1
$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04
$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04
add a comment |
1 Answer
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oldest
votes
$begingroup$
I guess you can discretize the raw price process too instead of the log price process. You get
$$
S_{t+1} = S_t + mu S_t + sqrt{v_t} S_t Z_t
$$
(where $Z_t$ is a standard normal variate), or
$$
frac{S_{t+1}}{S_t} - 1 = mu + sqrt{v_t} Z_t.
$$
Got the idea from: https://arxiv.org/pdf/1707.00899.pdf
answered Jan 12 at 5:30
TaylorTaylor
287113
$endgroup$
add a comment |
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$begingroup$
I guess you can discretize the raw price process too instead of the log price process. You get
$$
S_{t+1} = S_t + mu S_t + sqrt{v_t} S_t Z_t
$$
(where $Z_t$ is a standard normal variate), or
$$
frac{S_{t+1}}{S_t} - 1 = mu + sqrt{v_t} Z_t.
$$
Got the idea from: https://arxiv.org/pdf/1707.00899.pdf
answered Jan 12 at 5:30
TaylorTaylor
287113
$endgroup$
add a comment |
$begingroup$
I guess you can discretize the raw price process too instead of the log price process. You get
$$
S_{t+1} = S_t + mu S_t + sqrt{v_t} S_t Z_t
$$
(where $Z_t$ is a standard normal variate), or
$$
frac{S_{t+1}}{S_t} - 1 = mu + sqrt{v_t} Z_t.
$$
Got the idea from: https://arxiv.org/pdf/1707.00899.pdf
answered Jan 12 at 5:30
TaylorTaylor
287113
$endgroup$
add a comment |
$begingroup$
I guess you can discretize the raw price process too instead of the log price process. You get
$$
S_{t+1} = S_t + mu S_t + sqrt{v_t} S_t Z_t
$$
(where $Z_t$ is a standard normal variate), or
$$
frac{S_{t+1}}{S_t} - 1 = mu + sqrt{v_t} Z_t.
$$
Got the idea from: https://arxiv.org/pdf/1707.00899.pdf
answered Jan 12 at 5:30
TaylorTaylor
287113
$endgroup$
I guess you can discretize the raw price process too instead of the log price process. You get
$$
S_{t+1} = S_t + mu S_t + sqrt{v_t} S_t Z_t
$$
(where $Z_t$ is a standard normal variate), or
$$
frac{S_{t+1}}{S_t} - 1 = mu + sqrt{v_t} Z_t.
$$
Got the idea from: https://arxiv.org/pdf/1707.00899.pdf
answered Jan 12 at 5:30
TaylorTaylor
287113
answered Jan 12 at 5:30
TaylorTaylor
287113
answered Jan 12 at 5:30
TaylorTaylor
287113
answered Jan 12 at 5:30
TaylorTaylor
287113
287113
add a comment |
add a comment |
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$begingroup$
Be aware: crossposted.
$endgroup$
– Bob Jansen
Jul 11 '18 at 15:04