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frolloos
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Are these two RVs independent?

June 16th, 2018, 1:24 pm

Let

[$] Z_T = \exp \left( -\frac{1}{2} \int_t^T x_u^2 du + \int_t^T x_u dW_u \right ) [$]

and

[$] Y_T= E_T \left [ f \left( \int_T^{T'} x_u^2 du \right ) \right ] [$]

What is 

[$] E_t [Y_TZ_T] ? [$]

I think I can write

[$] E_t [ Y_T Z_T] = E_t [Z_T | Y_T] E_t [Y_T] [$]

Why would [$] Z_T [$], which depends on paths up to time [$] T [$], be dependent on [$] Y_T [$] which is the T-expectation of some function over *future* paths in the interval [$] [T,T'] [$]?

This is probably basic, but I'm confused.
 
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Alan
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Re: Are these two RVs independent?

June 16th, 2018, 3:38 pm

Not independent, as knowledge of one will change expectations of the other, so [$]Z_T[$] is affected by knowledge of [$]Y_T[$]. Think of Brownian bridges: knowledge that [$]B_T = a[$] affects path probabilities of [$]B_t[$] for [$]t < T[$]. 

I'm not sure about the relation below "I think I can write", but then that wasn't your question. Certainly that relation would be true if the E's were P's, as then it would be the defn of conditional probability.
 
frolloos
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Re: Are these two RVs independent?

June 16th, 2018, 5:05 pm

I'm almost sure you're right, but I'm still puzzled.

Why would [$]Y_T [$] depend on the *path* taken in the interval [$] [t,T] [$]? I can see that it will strongly depend on the value [$] x_T [$], but not why it should strongly depend on the say the whole path [$] \int_t^T x_u dW_u [$]. Maybe it depends on the path in a "weak" sense, although I don't know how to express strong or weak mathematically in this context.

Or from the "other side", I can intuitively understand that [$]Z_T [$] depends on the value [$] x_T [$] (which is I think what you mean with the Brownian bridge example), but can't see why it should depend on [$] x_u, u> T [$], and hence why it depends on [$] Y_T[$].
 
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Alan
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Re: Are these two RVs independent?

June 16th, 2018, 5:32 pm

Well, I am just saying [$]Z_T[$] and [$]Y_T[$] are not independent. At a minimum, they both depend upon [$]x_T[$] -- as you said -- which suffices for lack of independence. 

What is [$]\{x_t\}[$], anyway -- a diffusion independent of [$]W[$]?  
 
frolloos
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Re: Are these two RVs independent?

June 16th, 2018, 5:42 pm

Yes, in a calculation I did I too hastily assumed that 

[$] \epsilon E_t \left[ \left ( \int_t^T x_u dW_u \right) Y_T \right] \approx 0 [$]

which is basically an independence assumption (epsilon is a small parameter). I see now that it's not correct.

Thanks Alan.

No, x_u is driven by W_u, so dx_u = a(x_u,u) dW_u + drift term
 
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bearish
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Re: Are these two RVs independent?

June 17th, 2018, 2:19 pm

Hang on. I think you can get there by a slightly different conditioning argument: for any path of x, the stochastic integral will have mean zero, and therefore the product of it and Y will have mean zero. Integrating back over the distribution of x will preserve that zero mean. 
 
frolloos
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Re: Are these two RVs independent?

June 17th, 2018, 3:20 pm

Hang on. I think you can get there by a slightly different conditioning argument: for any path of x, the stochastic integral will have mean zero, and therefore the product of it and Y will have mean zero. Integrating back over the distribution of x will preserve that zero mean. 
Not sure what kind of conditioning argument will work here, can you say a bit more about your approach? So, the problem I am looking at is to calculate:

[$] \epsilon E_t \left[ \left (\int_t^T x_u dW_u \right ) E_T \left [ \sqrt{\int_T^{T'} x_u^2 du} \right ] \right ][$]

with [$] | \epsilon |< 1[$]. I don't need to explicitly calculate the expectation of the square root, if the whole thing can be shown to be approximately zero - which I doubt - then I'm a happy man :)
 
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bearish
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Re: Are these two RVs independent?

June 17th, 2018, 6:31 pm

Sorry - I somehow missed the point that x was driven by the same Brownian motion as it is being integrated against. In that case, the full path of x contains information about the future values of W, which renders my approach useless. I played around a bit with a simple binomial discretization, and it didn't look promising. 
 
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Edophokles
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Re: Are these two RVs independent?

June 18th, 2018, 6:26 am

What do you actually mean by E_T and E_t? Is it the conditional expectation given all the information generated by W_s  till the time T and t, respectively?
 
frolloos
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Re: Are these two RVs independent?

June 18th, 2018, 6:35 am

What do you actually mean by E_T and E_t? Is it the conditional expectation given all the information generated by W_s  till the time T and t, respectively?
Yes, that's right.
 
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Edophokles
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Re: Are these two RVs independent?

June 18th, 2018, 8:47 am

Okay, in that case \( Y_T \) can be dependent on the whole information till time \(T \) by definition. Here, some simple example, \(\int_1^2B_tdt\) is not independent of \(B_{0.5} \).
 
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Alan
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Re: Are these two RVs independent?

June 18th, 2018, 2:21 pm

I suspect everybody is agreed at this point that in this problem we have two random variables [$]A[$] and [$]B[$] that are not independent. However, just because [$]A[$] and [$]B[$] are dependent, does not mean [$]E[A B] = 0[$] cannot be true. For example, say [$]A[$] is a real random variable with a symmetric distribution about [$]A=0[$] and [$]B = A^2[$].

However, I suspect the expectation in this problem is indeed not zero. A somewhat simpler variation on it takes [$]dx_u = x_u dW_u[$] and [$]Y_T = x_T[$]. Then, the expectation in question (taking t=0 for simplicity) is 

[$]E_0[(\int_0^T x_t dW_t) x_T] = E_0 [(x_T - x_0) x_T] = E_0[x_T^2] - x_0^2 \not= 0[$]   

by elementary calculation. This example makes me think the more complicated expectation in the problem is not zero either.  (Perhaps one should look at [$]E_0 [(x_T - x_0) |x_T|][$] as being a little closer to the real example). 

As to whether it is non-zero, but small enough to neglect anyway in some larger calculation: impossible for third parties to say. How small it is depends on how large [$]T[$] is in my simpler example, and so probably how large [$]T' - T[$] is in the real example, as well as the other details of that case like the unspecified [$]a(x_u,u)[$] and drift. Once those unspecified things are given, a simple Monte Carlo could answer the question: "how small is it relative to other quantities of the same dimensions in the larger calculation?" 
 
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Edophokles
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Re: Are these two RVs independent?

June 18th, 2018, 3:41 pm

I think, unfortunately neither of the statements can be proved for general \(x_t\). Consider a trivial example of any deterministic process \(x_t\). Then the random variables \(Z_T\) and \(Y_T\) are independent (and hence the expectation of the product of these random variables becomes zero due to \(\mathbb{E}(Z_Z)=0\)). A non-trivial example would be \(x_t = B_t\) and \(Y_T = x_T\). In this case, the random variables are not  independent, but as mentioned in the post by Alan it is not a necessary condition for \(\mathbb{E}(Z_TY_T)=0\) to be true. 

For some particular \(x_t\) one would be able to compute the expectation of the product explicitly and see if the answer is close to zero or not.   
 
frolloos
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Re: Are these two RVs independent?

June 18th, 2018, 4:13 pm

As to whether it is non-zero, but small enough to neglect anyway in some larger calculation: impossible for third parties to say. How small it is depends on how large [$]T[$] is in my simpler example, and so probably how large [$]T' - T[$] is in the real example, as well as the other details of that case like the unspecified [$]a(x_u,u)[$] and drift. Once those unspecified things are given, a simple Monte Carlo could answer the question: "how small is it relative to other quantities of the same dimensions in the larger calculation?" 
The context of the problem is the following: for [$] T'-T [$] small enough we can write

[$] C(S_T, T, S_T, T', \sigma_T) \approx 0.4 S_T \sigma_T \sqrt{T'-T} [$]

For [$] T' - T [$] not so small, and choosing a slightly different forward strike, I can derive

[$] \text{Something I can observe today} \approx E_t  \left[ S_T E_T \left [ \sqrt{\int_T^{T'} \sigma^2_u du} \right ] \right ] [$]

Of course I can always work under the share measure, and then I'm done, i.e. I can relate the expected forward realized volatility ( the forward start volswap strike) to something I can observe today, but under the *share measure*. The problem is, I am not sure how to interpret the expected forward start realized volatility under the share measure  other than that the drift is different. Maybe I should look at the variance strike under the share measure to gain more feeling for it? But, if I insist on working under the risk-neutral measure, the independence question above pops up.

However, as in general there is no independence I think focusing on what expected future vol under the share measure means is maybe more useful.
 
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Alan
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Re: Are these two RVs independent?

June 18th, 2018, 6:58 pm

So, if you want calibrate or compare the r.h.s. to the l.h.s., it should be reasonably straightforward to calculate the r.h.s. numerically for any parameterized model of interest. Do that, and it seems you're done.