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Amin
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Re: Positive Heston European call theta

May 15th, 2018, 7:12 am

Just trying to understand on my own, if [$]t_2[$]>[$]t_1[$] and terminal time is set at T. If instantaneous variance is denoted as var(t)
[$]\int_{t_1}^T var(s) ds =\int_{t_1}^{t_2} var(s) ds + \int_{t_2}^T var(s) ds[$]

So any parameter in the SDEs can dramatically change at [$]t_2[$] or after [$]t_2[$] but still the value of 
[$] \int_{t_1}^T var(s) ds [$]
cannot be smaller than
[$]\int_{t_2}^T var(s) ds[$]
 
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lovenatalya
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Re: Positive Heston European call theta

May 15th, 2018, 7:27 am

Alan wrote:
lovenatalya wrote:
Oh, yes, Alan. You are absolutely right about the time-homogeneous case. I was distracted and blindsided by other factors like the convexity of the second partial derivative of the volatility and thus embarrassingly obtuse.

Anyway, I have the answer to the time-inhomogeneous case now. The answer is affirmative. Let us examine the following simple formulation

[$]\frac{dS}{S} = s(t)\sqrt{v(t)} dB_S[$]

[$]v(t)[$] is some time-homogeneous positive Ito process with its coefficients independent of [$]S[$]. [$]s(t)[$] is a time-dependent deterministic positive function. Now take the example of Heston with volatility of volatility equal to zero and constant positive [$]\lambda[$]

[$]dv = -\lambda(v-1)dt[$]

[$]v(t)[$] is a time-homogeneous process. The total variance
 
[$]\text{var}(t)=\int_t^T s(\tau)^2 v(\tau-t) d\tau.[$] 

So long as [$]v(t)[$] decreases fast (for the specific example of above, [$]v(t=0)>1[$] and [$]\lambda[$] is somewhat large), and [$]s(t)[$] increases fast enough, var[$](t)[$] increases with [$]t[$] and so does the call [$]C(t,S,\sigma)[$] at least for some [$]t[$]. In essence, the mechanism is simply the following. Given positive tuples [$](v_1,v_2)[$] and [$](s_1,s_2)[$] we want 

[$]v_1s_1+v_2s_2<v_1s_2\Longleftrightarrow \big( (\frac{v_2}{v_1}+\frac{s_1}{s_2}<1) \wedge (v_2<v_1) \wedge (s_1<s_2)\big).[$] 


In conclusion, theta of a zero rate European call is always nonpositive for a time-homogeneous process, but can be positive for a time-inhomogeneous process.

Your May 14 argument relies on this earlier argument that I have quoted here. But, IMO, this earlier argument is unconvincing as it stands. Using your expression above for var(t), please just give any specific example of [$](\lambda,v(0))[$] and the function [$]\{s(\tau): t \le \tau \le T\}[$] such that
 
[$] \frac{d}{dt} var(t) > 0[$]

and I will believe the whole thing.

Thanks.

You may view my previous example of discrete summation as the result of integration over a stepwise increasing function [$]s(t)[$]. There could be a trickier and cleaner example, but here is a construct for differentiable functions which I suppose is more to your taste.  Please check my computation.

Proof:  Let [$]s(t)^2=e^{kt}[$] and [$]v(t)-1 =(v_0-1) e^{-\lambda t}[$] for positive constants [$]k[$], [$]\lambda[$] and [$]T[$] to be determined. Substitute them into the var integral, we have 

[$]\text{var}(t) = \int_t^T s(\tau)^2v(\tau-t)\,d\tau= (v_0-1) e^{\lambda t}\int_t^T e^{(k-\lambda)\tau}\,d\tau+\int_t^T e^{k\tau}\,d\tau[$]

and
[$]\frac{\partial}{\partial t}\text{var}(t) = (v_0-1)e^{\lambda t}(\lambda\int_t^T e^{(k-\lambda)\tau}\,d\tau-e^{(k-\lambda)t})-e^{kt}[$]
[$]=\frac{\lambda}{k-\lambda}e^{kt}\big[(v_0-1)\big(e^{(k-\lambda)(T-t)}-\frac{k}{\lambda}\big)-\big(\frac k\lambda-1\big)\big].[$]

There are myriads of ways to make the above partial derivative positive. For example, we can set [$]k>\lambda>0[$], then make [$]v_0-1[$] and [$]T-t[$] all positive and large enough. 
 
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lovenatalya
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Re: Positive Heston European call theta

May 15th, 2018, 7:56 am

Amin wrote:
Just trying to understand on my own, if [$]t_2[$]>[$]t_1[$] and terminal time is set at T. If instantaneous variance is denoted as var(t)
[$]\int_{t_1}^T var(s) ds =\int_{t_1}^{t_2} var(s) ds + \int_{t_2}^T var(s) ds[$]

So any parameter in the SDEs can dramatically change at [$]t_2[$] or after [$]t_2[$] but still the value of 
[$] \int_{t_1}^T var(s) ds [$]
cannot be smaller than
[$]\int_{t_2}^T var(s) ds[$]

This is for a set volatility path. What if the whole path is changed when the starting time shifts?
 
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Alan
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Re: Positive Heston European call theta

May 15th, 2018, 2:49 pm

lovenatalya wrote:
[$]\text{var}(t) = \int_t^T s(\tau)^2v(\tau-t)\,d\tau= (v_0-1) e^{\lambda t}\int_t^T e^{(k-\lambda)\tau}\,d\tau+\int_t^T e^{k\tau}\,d\tau[$]

and
[$]\frac{\partial}{\partial t}\text{var}(t) = (v_0-1)e^{\lambda t}(\lambda\int_t^T e^{(k-\lambda)\tau}\,d\tau-e^{(k-\lambda)t})-e^{kt}[$]
[$]=\frac{\lambda}{k-\lambda}e^{kt}\big[(v_0-1)\big(e^{(k-\lambda)(T-t)}-\frac{k}{\lambda}\big)-\big(\frac k\lambda-1\big)\big].[$]

There are myriads of ways to make the above partial derivative positive. For example, we can set [$]k>\lambda>0[$], then make [$]v_0-1[$] and [$]T-t[$] all positive and large enough. 

Very good -- I'm convinced, as it seems likely a small [$]\eta  > 0[$] would not wreck the idea. (I still would argue for for my previous "total vol derivative" (***) for theta when [$]\eta=0[$]). 

So, are there any financial implications of having a positive theta from this mechanism (we already saw it could be achieved by the cost-of-carrys), or is it just an interesting mathematical curiosity?
 
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lovenatalya
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Re: Positive Heston European call theta

May 16th, 2018, 5:34 pm

Alan wrote:
Very good -- I'm convinced, as it seems likely a small [$]\eta  > 0[$] would not wreck the idea. (I still would argue for for my previous "total vol derivative" (***) for theta when [$]\eta=0[$]). 

The positive [$]\eta[$] does not have to be small make the conclusion true, so long as the [$]s(t)[$] increases fast and [$]v(t)[$] decreases fast as you can see from the proof. For the deterministic case, I agree that the total derivative (***) should be used since there is only one predetermined volatility curve.

Alan wrote:
So, are there any financial implications of having a positive theta from this mechanism (we already saw it could be achieved by the cost-of-carrys), or is it just an interesting mathematical curiosity?

This comes from a historical investigation of a long-dated European option under some stochastic volatility process similar to the Heston model with time-dependent coefficients and zero interest/dividend rates. I was computing the time difference of the Monte Carlo simulation price and was surprised to find some positive numbers. I tried to see if the positivity was an indication of a numerical error or admissible mathematically. Now we know the latter is true.


Here is a further question. I would like to use the mixing formula to compute theta by Monte Carlo simulation. We know that for a European call [$]C[$], the price is

[$]<C_{\text{BS}}(S_{\text{eff}},V_{\text{eff}})>[$]

where [$]C_{\text{BS}}[$] stands for the Black-Scholes call formula, [$]<\cdot>[$] denotes the expectation with respect to the Brownian motion of the stochastic volatility process and [$]\cdot_{\text{eff}}[$] indicate the effective quantity conditioned on a stochastic volatility Brownian motion path. Can we derive a mixing formula for the theta of [$]C[$] in the form of

[$]<\frac{\partial}{\partial S}C_{\text{BS}}(S_{\text{eff}},V_{\text{eff}})>+<f>[$]

for some function [$]f[$], similar to the two terms in [$]U_{t,1}[$] in my previous post time stamped Wed May 09, 2018 12:13 am GMT?
Last edited by lovenatalya on May 16th, 2018, 8:03 pm
 
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Alan
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Re: Positive Heston European call theta

May 16th, 2018, 5:55 pm

Gee, on the last question, I have no idea -- seems like an issue for you, not me.
 
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lovenatalya
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Re: Positive Heston European call theta

May 16th, 2018, 8:00 pm

Alan wrote:
Gee, on the last question, I have no idea -- seems like an issue for you, not me.

Oh yes, it is, of course, my issue. :-)  But I would like to pick your awesome brain to find a solution. The issue is, as we all know, numerical differentiation is unstable and prone to noise. I would like to use the mixing formula to analytically perform all differentiations and leave only the summation (averaging) to the numerical (Monte Carlo) method. I think the mixing formula is a natural way to do this. Does this problem interest you? Do you think it is better to open another thread for this problem?
 
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Alan
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Re: Positive Heston European call theta

May 17th, 2018, 2:17 pm

OK, I got interested enough to think about it a little. This thread is fine. Let's see if we can agree on how to do it for [$]\rho=0[$] and time-homogeneous models. Start from

(M1) [$]C(t,T,S_t,V_t) = <c_{BS}(t,T,S_t,V_{eff}(t,T,V_t)>[$],

where the brackets denote V-path averaging. The expression inside the brackets is now the BS volatility under (pathwise) deterministic volatility, and I believe you have come around to my way of thinking about how to do theta for that. So, my first thought is that theta, in this case, should be somethiing like the bracket average of my earlier (*) where the May 8 [$]U(t,T,V_t) = (T-t) V_{eff}(t,T,V_t)[$]. Agree?

If so, the way I would work is to confirm this guess by some numerics, say the Heston model again. But I am not so interested to do that here, so I leave that up to you. 
 
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lovenatalya
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Re: Positive Heston European call theta

May 18th, 2018, 8:24 am

For time-homogeneous models, we do not have any disagreement. In fact your conjecture is just the first term of mine in the post on Wed May 16, 2018 5:34 pm GMT
[$]\big\langle\frac{\partial}{\partial S}C_{\text{BS}}(S_{\text{eff}},V_{\text{eff}})\big\rangle+\langle f\rangle.[$]

Indeed, the conjecture I had in mind when I wrote that post was theta is the volatility path bracket expectation of your (*) on Tue May 08, 2018 8:49 pm GMT with [$]U_t[$] replaced by

[$] U_{t,1}(t_0) = -\beta(t_0,t_0,\beta_0)^2+\int_{t_0}^T\frac{\partial}{\partial t_0}\beta(s,t_0,\beta_0)^2\,ds,[$]

as I have defined in my post Wed May 09, 2018 12:13 am GMT. The first term is your term for the time homogeneous case and the second additional term is [$]f[$] for when it is time-inhomogeneous. I am trying to prove this, first for the time-homogeneous case, i.e. the first term only, then for the time-inhomogenous one, i.e. together with the second term. It does not seem trivial though.
 
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Alan
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Re: Positive Heston European call theta

May 18th, 2018, 2:21 pm

Well, you sucked me in -- clever boy (or girl).  :D

I think the [$]\beta[$] notation is a distraction that overly complicates things. 

If we simply write

[$] dS_t = \sqrt{V_t} S_t dB_t[$], where [$]V_t[$] is some diffusion (time-homogeneous or not, driven by a Brownian uncorrelated to [$]B_t[$]), then it seems to me that 

(M2) [$]C_t(t,T,S_t,V_t) =  \left< \frac{S_t \, e^{-d_1^2/2}}{2 \sqrt{2 \pi \, U(t,T)}} \, U_t \right> = - S_t V_t  \left< \frac{e^{-d_1^2/2}}{2 \sqrt{2 \pi \, U(t,T)}} \right>[$],

since [$]U(t,T) = \int_t^T V_s \, ds[$].

Proof: just differentiate [$]C_{BS}[$] w.r.t. t inside the brackets, so no different than a proof of my earlier (*) under deterministic volatility.

To implement (M2), you just simulate many V-path's, all starting at [$]t[$] with the same [$]V_t[$], evaluate [$]U(t,T)[$] for each such path, and take the MC average of the final expression in brackets. It's not much different than using mixing to evaluate [$]C[$] itself, where you do the same thing but in that case the "expression in brackets" is the BS formula. So, each MC run can yield both [$]C[$] and [$]C_t[$] with little additional work, as you have to evaluate [$]U(t,T)[$] and [$]d_1[$] for the BS formula anyway.

Again, if doubtful, I'd say try it -- but I have noticed you are reluctant for some reason to post numerical examples.
 
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lovenatalya
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Re: Positive Heston European call theta

May 18th, 2018, 5:28 pm

Alan wrote:
Well, you sucked me in -- clever boy (or girl).  :D

Hahahaa, I am rather glad. It is the problem itself that is intriguing. :-D

 Proof: just differentiate [$]C_{BS}[$] w.r.t. t inside the brackets, so no different than a proof of my earlier (*) under deterministic volatility.

I do not think the reasoning here is sufficient. This is actually the deceiving part. This is how I obtained my conjecture. However, there are two layers of integrations both of which depend on [$]t[$] and [$]T[$]. We shift [$]t[$], the SDE for the volatility shifts as well, affecting the Black-Scholes integral in turn. We can not blithely differentiate only one integration and disregard the other, at least not without proper justification. We conjecture/wish our bold and cavalier claim is right. But that is no proof. For the time-homogeneous case, it is easier. We can differentiate instead [$]T[$]. However, that tail end still moves the volatility SDE just the same, even if it is just the tail end. For the time-inhomogeneous case with coefficients all time-dependent, there is no way around the [$]t[$]-shift of the whole volatility SDE. The whole path of the volatility needs to be looked at. It may turn out that the differentiation on the BS integral is all that is needed as you and I have conjectured/wished, but that needs rigorous justification. The proof is not so simple.

I think I am close to getting the time-homogeneous case but need some more time to see if the approach works. 

To implement (M2), you just simulate many V-path's, all starting at [$]t[$] with the same [$]V_t[$], evaluate [$]U(t,T)[$] for each such path, and take the MC average of the final expression in brackets. It's not much different than using mixing to evaluate [$]C[$] itself, where you do the same thing but in that case the "expression in brackets" is the BS formula. So, each MC run can yield both [$]C[$] and [$]C_t[$] with little additional work, as you have to evaluate [$]U(t,T)[$] and [$]d_1[$] for the BS formula anyway.

Again, if doubtful, I'd say try it -- but I have noticed you are reluctant for some reason to post numerical examples.

I know how to do the Monte Carlo simulation and implement the mixing formula. The whole problem itself stems from this very operation. However, the very reason to seek a mixing formula for the time differentiation of the call option is that numerical differentiation is suspect, and thus not to be trusted. The time difference of the simulated option value looks noisy/oscillatory for short time difference. The level of oscillation depends on the number of paths as well. Yes, we can compute easily our conjectured formulae using the mixing formula. But we can not use the Monte Carlo time derivative as the benchmark. One thing we can do, but only for the particular case of the Heston model, is to use the semi-analytic Fourier transform formula of Heston to compute the time derivative as the benchmark. However, I would first spend some time to see if my theoretical approach works for the proof of the homogeneous case because it looks promising.
 
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Alan
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Re: Positive Heston European call theta

May 19th, 2018, 2:18 pm

Fine. After you do that, let's see if you still believe my proposed mixing formula (M2) for [$]C_t[$] is wrong -- even for the time-homogeneous case, such as standard Heston model. You could be right, but then please post a numerical example where you demonstrate this. (We are both agreed that one can be done two ways).    

May I suggest sticking to the Heston parameter set I already gave as an example (May 11), except now [$]\eta = 1[$]. That way, you won't have to post anew the parameters. Plus, if you use those, I will likely have time to check your numerics, and progress will be made.
 
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lovenatalya
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Re: Positive Heston European call theta

May 19th, 2018, 7:57 pm

Alan wrote:
Fine. After you do that, let's see if you still believe my proposed mixing formula (M2) for [$]C_t[$] is wrong -- even for the time-homogeneous case, such as standard Heston model. You could be right, but then please post a numerical example where you demonstrate this. (We are both agreed that one can be done two ways).    

You have misunderstood my last comment. I said your (M2) was a special case of my conjectured mixing formula when the parameters were all constant or time-homogeneous. I know how you got your formula because that was exactly how I got my more general one. However, the proof is not complete because it boldly/cavalierly ignores crucial issues that need to be addressed. This disregard for rigor was how I obtained my conjecture. That is fine for proposing a conjecture. But this disregard disqualifies it from being a proof. In short, I agree with your conjecture but disagree with your proof.

May I suggest sticking to the Heston parameter set I already gave as an example (May 11), except now [$]\eta = 1[$]. That way, you won't have to post anew the parameters. Plus, if you use those, I will likely have time to check your numerics, and progress will be made.

Sure. The numerics check could only be for the constant coefficient Heston model and not for anything else, though.
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