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Alan
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Re: Positive Heston European call theta

May 9th, 2018, 1:21 am

1. I agree they are different

2. I agree with the last equalities in both lines.

3. I don't like the definitional equalities as it suggests there are two ways to compute [$]U_t[$] and there really isn't. You will probably say that I can't complain about definitions, but I can when we are talking about [$]U_t[$]. As I keep repeating, [$]t_0[$] is fixed.  

I'd say, if you want to go one step here, let's indeed. Take the example [$]\beta = 2 \, e^{-t}[$] which was your example with [$]t_0=0[$]. Next, do the integral for [$]U = \int_t^1 \beta^2(s) \, ds[$]. Now compute [$]U_t[$]. I know you agree it's line 2. End of story, IMO.  
 
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lovenatalya
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Re: Positive Heston European call theta

May 9th, 2018, 4:36 am

Let me first write down the explicit answer to your last paragraph regarding my simple example. Let's keep [$]t_0[$] instead of setting it to [$]0[$] so as to make clear its dependence on [$]t_0[$]. The beta function is [$]\beta(t,t_0,\beta_0)=\beta_0 e^{-(t-t_0)}[$]. For line 1, [$]U_{t,1}=-\beta_0^2 e^{-2(T-t_0)}[$]. For line 2, [$]U_{t,2}=-\beta_0^2[$]. Do you agree?

Let me continue on with the inquiry. Let us start from the beginning.

We value two calls with the same stock price [$]S_0[$] and the same volatility [$]\beta_0[$] at two distinct valuation times [$]t_1[$] and [$]t_2[$]. We denote the two call values as [$]C(t_1,\beta_0)[$] and [$]C(t_2,\beta_0)[$]. 

The volatility process obeys one ordinary differential equation

[$]d\beta = f(t,\beta)dt[$]

for a uniformly Lipschitz continuous function [$]f[$]. There is a family of solutions [$]\beta(t,t_0,\beta_0)[$] such that for any given [$](t_0,\beta_0)[$], [$]\beta(t,t_0,\beta_0)[$] is the unique solution satisfying the aforementioned ODE in the variable [$]t[$] with its trajectory going through [$](t_0,\beta_0)[$], i.e., [$]\beta(t=t_0,t_0,\beta_0)=\beta_0[$].

For the two call options described in the first paragraph, the volatility function has to satisfy the ODE and the two distinct initial conditions. That means [$]\beta(t,t_1,\beta_0)[$] where [$]\beta(t=t_1,t_1,\beta_0)=\beta_0[$] is the volatility function for the first call and [$]\beta(t=t_2,t_2,\beta_0)=\beta_0[$] is the volatility function for the second call. These two functions describe two different trajectories. In other words, [$]\beta(t,t_1,\beta_0)[$] and [$]\beta(t,t_2,\beta_0)[$] are two different functions in the variable [$]t[$]. Otherwise, [$]\beta(t_1,t_1,\beta_0)=\beta(t_1,t_2,\beta_0)=\beta(t_2,t_1,\beta_0)=\beta(t_2,t_2,\beta_0)=\beta_0[$] which is generally false.

Do we agree thus far?
 
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Alan
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Re: Positive Heston European call theta

May 9th, 2018, 1:52 pm

Not at all. In your last paragraph starting with 'For ..', you are thinking about it wrong. There are *not* "two distinct initial conditions". When you say earlier that "We value two calls with the same stock price [$]S_0[$] and the same volatility [$]\beta_0[$] at two distinct valuation times [$]t_1[$] and [$]t_2[$]", that is not correct thinking about deterministic volatility, as I will explain.  (The 'same stock price' part is correct; the 'same volatility' part is where you are confused).

There are three times: [$](t_0,t_1,t_2)[$].

[$]t_0[$] is a fixed time which you have introduced to define your deterministic volatility function. That's fine. But, this time and this function must be fixed, so also [$]\beta_0[$] is fixed. 

Imagine drawing the trajectory on a piece of paper: the horizontal axis is the time and the vertical axis is the instantaneous volatility [$]\beta[$] at that time. Make a mark on the time axis for [$]t_0[$]. Above that mark, the volatility function value is [$]\beta_0[$]. 

Now make two more marks on the time axis at [$]t_1[$] and [$]t_2[$]. These are the option valuation times. As [$]t_2 \rightarrow t_1[$], you can deduce [$]C_t[$] at [$]t = t_1[$]. These times [$](t_1,t_2)[$] play no role in the initial conditions for the volatility (as you have incorrectly asserted).  You must think of the volatility function as a property of the stock. It exists whether or not options on the stock exist. It can be measured (at each time t) by observations on the stock price trajectory. The introduction of two options does not change this postulated property (the instantaneous volatility) of the stock price process. Thus, the introduction of two arbitrary option valuation times [$](t_1,t_2)[$] does not change the volatility process; i.e. the volatility trajectory function.

That's the correct way to think about it. 

To summarize: there aren't two volatility trajectories; there's only one trajectory, the one you just drew on your piece of paper.

As I suggested, fix a volatility function and run some Monte Carlo's if you aren't convinced. Or go look up Merton or somebody that has formulas for option valuation under deterministic volatility: I am not inventing a new concept here. (They are in my first book on pg 65, but you may want a third party source). The formulas I wrote for [$]C(t,T,S_t)[$] are standard. If you agree with them, then apparently we are arguing about how to take a partial derivative. If you don't agree with them, then what do you think is the formula for a call value at time [$]t < T[$], where [$]T[$] is a fixed expiration, and where the instantaneous volatility is [$]\beta(s) = 2 \, e^{-s}[$], [$](0 \le s < \infty)[$], the example I adapted from yours? I think we need to have agreement on that to make any further progress if you are still doubtful at this point.
 
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lovenatalya
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Re: Positive Heston European call theta

May 9th, 2018, 8:42 pm

I want to clarify first that, I am not arguing against the essentially Black-Scholes formula with the time-dependent volatility per se, you wrote in your previous post, as evidenced by my ready agreement before. Our difference is how to take the partial derivative of the variance integral [$]U[$] with respect to time, specifically whether it should be [$]U_{t,1}[$] (line 1) or [$]U_{t,2}[$] (line 2).

You said:
The formulas I wrote for [$]C(t,T,S_t)[$] are standard. If you agree with them, then apparently we are arguing about how to take a partial derivative.
So, yes, we are indeed arguing about how to take a partial derivative.
 
I understand your reasoning regarding the deterministic volatility case. I thought of the same thing before I came to my current and different conclusion. I think the confusing factor is that we are now talking about the deterministic volatility as an end to itself. Yes, I used the deterministic volatility, but only to make a simplified computation as an extreme and degenerate example of the stochastic volatility case, not as an end to itself. My ultimate goal is to investigate the proper stochastic volatility, not the deterministic volatility. At this stage, this degenerate case seems to add unnecessary confusion. In this respect, the exponential example will not clarify but only highlight our difference, because my answer will be different from yours. I have already written out the two different partial derivatives, namely [$]U_{t,1}[$] and [$]U_{t,2}[$] for this exponential volatility function, in my last post. You want to pick [$]U_{t,2}[$], and I want to pick [$]U_{t,1}[$]. We already know our difference, particularly for this example. We now need to resolve it.
 
With the above preamble, let us back up and look at a proper stochastic volatility model instead of its degenerate deterministic volatility model. 

I understand you do not agree with setting the volatility arbitrarily, specifically to [$]\beta_0[$], at [$]t_1[$] and [$]t_2[$] for the deterministic volatility.  Now, for the proper stochastic volatility case, e.g., a Heston model with a positive volatility of volatility, do you agree we can? Specifically, we value two calls with the same stock price [$]S_0[$] and the same volatility [$]\beta_0[$] at two distinct valuation times [$]t_1[$] and [$]t_2[$]. We denote the two call values as [$]C(t_1,\beta_0)[$] and [$]C(t_2,\beta_0)[$]. Do you agree we can do that?
 
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Alan
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Re: Positive Heston European call theta

May 10th, 2018, 1:42 am

Yes, agree: not a problem there, assuming a so-called regular diffusion where the realized volatility state can be any positive number at every time [$]t[$]. For the Heston model, this is guaranteed by a strictly positive vol-of-vol.

Indeed, it's not only allowable but mandatory in that case to fix both the stock price and the volatility for purposes of computing what we have been calling [$]C_t[$]. In that case (a regular volatility diffusion), then the option value function is [$]C(t,S_t,V_t)[$] and so a notation I like is:

(**) [$]C_t  \equiv \left( \frac{\partial C}{\partial t} \right)_{S_t,V_t}[$]

which explicitly shows which variables are being held constant.

So, if you are evaluating [$]C_t[$] at [$]t=0[$], then you are fixing both [$]S_0[$] and [$]V_0[$] as time infinitesimally advances, which is what you asked about. 

I think this really clarifies our dispute above, because now I see you were arguing for the validity of (**) in the deterministic vol case.
 
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lovenatalya
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Re: Positive Heston European call theta

May 10th, 2018, 4:23 am

I think this really clarifies our dispute above, because now I see you were arguing for the validity of (**) in the deterministic vol case.
Excellent. Yes, (**) is what I am getting at. Allow me to attempt a leap and see if you agree. Does "I see ... " imply you now agree with the validity of (**) applied to the deterministic vol case, considered as a degenerate case of the stochastic vol model as volatility of volatility uniformly approaches zero and the fact that the call value is continuous with respect to the volatility of the volatility which in the Heston model is represented by a single number?
 
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Alan
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Re: Positive Heston European call theta

May 10th, 2018, 5:06 am

No. I still believe in what I said before. But I suspected you would argue, well, what if the Heston vol-of-vol is 10^(-16)? First, I"m not sure (**) as [$]\eta \downarrow 0^+[$] agrees with the "correct" [$]C_t[$] at [$]\eta = 0[$] (The correct value being the approach I argued for before). If it does, fine. If it doesn't, then the correct [$]C_t[$] is simply not continuous at [$]\eta = 0[$]. At the moment, I am agnostic about whether or not there is a discontinuity. 

It would be interesting to know if, for the Heston model, [$]C_t(\eta = 0^+) = C_t(\eta =0)[$], where the l.h.s is computed from (**) and the r.h.s is computed from a formula I labelled (*) in a May 8, 1:49pm post. Maybe I will compute an example at some point or you might want to.

In any event, remember that I already proved that [$]C_t \le 0[$] both in the Heston model and in what I consider to be the correct approach to the deterministic case. So, both of these [$]C_t's[$] are going to be negative.  What do you think -- are they going to agree?  
 
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lovenatalya
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Re: Positive Heston European call theta

May 10th, 2018, 6:28 am

Alright, I understand your position. I just want to make sure that my upcoming derivation is not preaching to the choir, so to speak. It is late, so I will write my derivation tomorrow. 

Also, when I say Heston model, I am including and particularly referring to the time-dependent coefficient Heston model. When you say
remember that I already proved that [$]C_t\le 0[$] ... in the Heston model 
you are referring to the constant coefficient Heston model, right? You have proved [$]C_t\le 0[$] for the time homogeneous model but not for the time-inhomogeneous one.
 
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Alan
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Re: Positive Heston European call theta

May 10th, 2018, 1:41 pm

Right, constant coefs.
 
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Alan
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Re: Positive Heston European call theta

May 11th, 2018, 2:33 pm

OK, I did some checks, and here I believe is a 'total derivative' relation that you will agree with. It makes explicit the difference between my preferred [$]C_t[$] and your preferred [$]C_t[$] for the deterministic vol. case. (Kind of my version of your [$]U_{t1}[$] and [$]U_{t2}[$]). Namely:

(***) [$] \left ( \frac{dC}{dt} \right)_{S_t} =   \left ( \frac{\partial C}{\partial t} \right)_{S_t,V_t} +   \left ( \frac{\partial C}{\partial V_t} \right)_{t,S_t} V'(t) [$].

I am using my previous notation that explicitly shows which variables are being held constant. The l.h.s. is simply somewhat new notation for my relation (*) that I posted earlier. I argued that (*) was the correct [$]C_t[$] under deterministic vol. You don't have to agree with the interpretational part, but I hope the definition as the previous formula (*) is clear.  

The first term on the r.h.s. is my previous (**), which I believe was your preferred expression for [$]C_t[$] under det. vol. Let's agree to evaluate it at [$]\eta = 0^+[$], an infinitesimally positive [$]\eta[$]. The second term on the r.h.s. is the "correction" term that makes (***) an equality. Again the r.h.s. partials all make sense at  [$]\eta = 0^+[$], and, separately, [$]V'(t)[$] makes sense at [$]\eta = 0[$], the deterministic limit.

Finally, for a numerical example, I took [$]S_0 = K =100[$], [$]V_0 = 1/2[$], [$]dV_t = (a - b V_t) dt + \eta \sqrt{V_t} dW_t[$], with [$]a = 1, b = 4, \eta = 1/1000[$], [$]T=1/4[$]. 

Numerically, I found for (***), at [$]t=0[$] with [$]V'(0) = a - b V(0) = -1[$]:

[$]   \left ( \frac{dC}{dt} \right)_{S_t}  = -30.8316[$]  (using the expression (*) that I posted earlier).

 [$] \left ( \frac{\partial C}{\partial t} \right)_{S_t,V_t} = -21.087[$] (using my Heston model code with the above [$]\eta[$] and a numerical t-derivative).

 [$] \left ( \frac{\partial C}{\partial V_t} \right)_{t,S_t} V'(0) = -9.7446[$] (again using my Heston model code with the above [$]\eta[$] and a numerical V-derivative).

Finally,

[$] -30.8316 = -21.087 - 9.7446 [$].

so (***) is checked and everything IMO makes sense. 

========================================

p.s. I am used to calling something like (***)  a 'total derivative' but, interestingly, wikipedia points out that there are eleven names for it! Since the stock price is being held constant, maybe something like 'total derivative with respect to the vol. flow only' would be a better name for it in this context.
 
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Re: Positive Heston European call theta

May 11th, 2018, 9:50 pm

I agree completely with the relationship you wrote. Yes, it is exactly the relation between the Lagrangian (total) derivate and the Eulerian (**) derivative in fluid dynamics.  I have not had time to write my proof of the existence of positive (**) or the Eulerian derivative for the time-dependent Heston model yet, but will do so either later today or tomorrow. Sorry for the delay.
 
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Re: Positive Heston European call theta

May 14th, 2018, 5:14 pm

Proof of the existence of positive European call theta for the Heston model with time-dependent coefficients and any other similarly structured time-inhomogeneous stochastic volatility models so long as the regularity of the call value PDE is assured.

Consider the Heston model. Let [$]C(t,\eta)[$] be the European call price evaluated at $t$ with stock price $S_0$ and variance [$]v_0[$] and a constant volatility of volatility [$]\eta[$] both expiring at the same time. In the derivation that follows, all parameters not appearing are assumed to be held constant.

We adopt the Andersen-Andreason version of the Heston model. The specific version is immaterial. It is only for convenience of derivation.

[$]\frac{dS}S = s(t)\sqrt v dW_S[$]
[$]dv = -\lambda(v-1)dt+\eta\sqrt v\, dW_v[$]

where [$]s(t)[$] is a postive deterministic function of [$]t[$]. I have already shown before, in post Mon May 07, 2018 11:03 pm GMT, that there is an increasing fast enough function [$]s(t)[$], a large enough [$]\lambda[$], two times [$]t_1<t_2[$] and the initial condition of the variance [$]v_0>1[$], such that [$]C(t_1,0)<C(t_2,0)[$]. (For [$]\eta=0[$], we take the mathematical interpretation of the Eulerian partial derivative with respect to [$]t[$], if you happen not to like the financial interpretation of it.) It can be proved that from the PDE of [$]C(t,\eta)[$], [$]C(t,\eta)[$] is continuous with respect to [$]\eta[$]. The proof for the general case is not trivial. So [$]\exists\eta_0>0\ni \big(0<\eta<\eta_0 \implies C(t_1,\eta)<C(t_2,\eta)\big)[$]. By Rolle's theorem, 
[$]\forall\eta\in(0,\eta_0),\,\exists t_0(\eta)\in[t_1,t_2]\ni \frac{\partial}{\partial t}C(t(\eta),\eta)>0[$].

[$]QED[$]
Last edited by lovenatalya on May 15th, 2018, 4:32 pm, edited 4 times in total.
 
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Amin
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Re: Positive Heston European call theta

May 14th, 2018, 10:50 pm

You may have some subtle point but as I understand that option prices without dividends and interest rates only depend upon integral of  variance which only decreases with time. 
You think life is a secret, Life is only love of flying, It has seen many ups and downs, But it likes travel more than the destination. Allama Iqbal
 
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lovenatalya
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Re: Positive Heston European call theta

May 15th, 2018, 12:23 am

You may have some subtle point but as I understand that option prices without dividends and interest rates only depend upon integral of  variance which only decreases with time. 
Please check your underlying process, your precise assertion, and your proof if you have one. You can compare it against and check my precise assertion and proof. Please let me know if you find any error.
 
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Alan
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Re: Positive Heston European call theta

May 15th, 2018, 4:06 am

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.