Clear explanation at end of video on how HJM can be made markovian as in Cheyette model. I think this is left between lines as "obvious" elsewhere and so this was very helpful.
Great video as always. One suggestion if I may of course: it would be very helpful if you can add subtitles for the non-native english speakers (like myself). Thank you for the videos again
Great Video, Loved it. I have a doubt though. Why at the time of proving f(t,T) is markov or not, you took D(t) = integral of [sigma(s,t)] with limits 0 to t, instead of D(t) = integral of [sigma(s,T)] with limits 0 to t. The second way of D(t) should have been used because in f(t,T) we have sigma(s,T)dWs term and not sigma(s,t)dWs. (timestamp: after 16:00). Please explain.
thanks for the comment! This is because we are assuming separable volatility - \sigma(t,T)=g(t)h(T) - so this (shall we say 'indirect' step) just helps us link the two arguments together.
I think Silvery's remark is solid and f(t,T) is markov if sigma_f(s,T) is deterministic. I think f(t,T) is markov even if sigma_f(s,T) depends on f(s,T).
Hello again, 18:24, we want to show that under log normal assumption(vol is const) there is arbitrage opportunity. Why we can discard the stochastic term, only keep the deterministic term please ?
Hi, very nice video. Just one question. At 5:30 you are calculating the sigma_P as an integral of sigma_f. Thus, you are integrating something positive and multiplying it by -1 to get something negative, the sigma_P. Is my understanding correct? Is it possible to get a negative sigma? Thanks
Great video, very intuitive explanation. One question though: where does Musiela parameterization come into picture and how does it simplify things for modeling HJM? Thanks.
Thanks Stephen R.! Musiela parameterisation is simple modification, and we have implicitly covered it in the first video (introductory video). We have written the instantaneous forward as f(t, T), where t is moving and T is fixed. Musiela writes the second argument in terms of remaining maturity, x=T-t, so if we substitute for T in our equation, we will get f(t,T)=f(t,t+x), so it is a function of t and x now, and Musiela represents this function as r(t,x), which is fair enough because the function spec has changed so renaming it make sense, and it is a function of t and x, so you can write (t,x) in place of (t, t+x) where the second t is redundant. This form makes it more obvious that f is infinite dimensional (think infinite space of functions). But we know in our representation there is one equation for each fixed T, so we already know it is an infinite dimensional problem.
Is the reason that we change risk neutral measure to t forward measure because the f under risk neutral measure has dt term, so it is not a martingale?
correct, plus also certain calculations might be easier/more natural under one measure than another, so having the transformation between the various measures comes in handy.
Hi, 16:30 D(T) - D(t) is decomposed into two terms, and you mentioned the first expectation is 0, but the second is not. Does that means if the expectation of D(T)-D(t) is zero then this process is markovian ? Why 0 expectation can imply markovian property ? Thanks!
No, if the expectation is zero it does not mean the process is Markovian (only that it is a Martingale) because we are only considering the first moment of the conditional distribution. However, since the first moment of D_T - D_t depends on the realisations of the Brownian up to and including time t, which are F_t-measurable, this is sufficient to conclude that the process is non-Markovian.
@@hihihi82 sigma_f(s,T) is deterministic, as opposed stochastic or constant, because it depends only on time. We are not observing at T - the equations are conditional on the filtration at time t < T. Deterministic does not mean measurable / observable, it just that we can perfectly predict what the value of the function will be because it depends only on time.
yes implicitly does boil down to Leiniz integral rule. We went in reverse order though - from derivative to integral. The two sub components are coming from the left hand side of the blue box - equations at the top and bottom. thanks!
20min here helped more than half a semester with my prof... good work!
Clear explanation at end of video on how HJM can be made markovian as in Cheyette model. I think this is left between lines as "obvious" elsewhere and so this was very helpful.
thank you!! much appreciated!
1 hour video kills my whole semester lol ! thanks a lot!
absolutely fantastic and very lucid video. thank you
Glad you enjoyed it! Very kind of you!
This is a topic I thought I would never understand, but I think I've finally get it for first time. Amazing stuff!
Great video as always.
One suggestion if I may of course: it would be very helpful if you can add subtitles for the non-native english speakers (like myself).
Thank you for the videos again
Great suggestion! We do for some, but should be more consistent!
Great Video, Loved it. I have a doubt though. Why at the time of proving f(t,T) is markov or not, you took D(t) = integral of [sigma(s,t)] with limits 0 to t, instead of D(t) = integral of [sigma(s,T)] with limits 0 to t. The second way of D(t) should have been used because in f(t,T) we have sigma(s,T)dWs term and not sigma(s,t)dWs. (timestamp: after 16:00). Please explain.
thanks for the comment! This is because we are assuming separable volatility - \sigma(t,T)=g(t)h(T) - so this (shall we say 'indirect' step) just helps us link the two arguments together.
I think Silvery's remark is solid and f(t,T) is markov if sigma_f(s,T) is deterministic. I think f(t,T) is markov even if sigma_f(s,T) depends on f(s,T).
Your content is great! Thank you very much for great explanation. Are you maybe planning to create videos about SABR model?
Many thanks! Indeed on the list!
Hello again, 18:24, we want to show that under log normal assumption(vol is const) there is arbitrage opportunity. Why we can discard the stochastic term, only keep the deterministic term please ?
Hi, very nice video. Just one question. At 5:30 you are calculating the sigma_P as an integral of sigma_f. Thus, you are integrating something positive and multiplying it by -1 to get something negative, the sigma_P. Is my understanding correct? Is it possible to get a negative sigma? Thanks
Great video, very intuitive explanation. One question though: where does Musiela parameterization come into picture and how does it simplify things for modeling HJM? Thanks.
Thanks Stephen R.! Musiela parameterisation is simple modification, and we have implicitly covered it in the first video (introductory video). We have written the instantaneous forward as f(t, T), where t is moving and T is fixed. Musiela writes the second argument in terms of remaining maturity, x=T-t, so if we substitute for T in our equation, we will get f(t,T)=f(t,t+x), so it is a function of t and x now, and Musiela represents this function as r(t,x), which is fair enough because the function spec has changed so renaming it make sense, and it is a function of t and x, so you can write (t,x) in place of (t, t+x) where the second t is redundant. This form makes it more obvious that f is infinite dimensional (think infinite space of functions). But we know in our representation there is one equation for each fixed T, so we already know it is an infinite dimensional problem.
Is the reason that we change risk neutral measure to t forward measure because the f under risk neutral measure has dt term, so it is not a martingale?
correct, plus also certain calculations might be easier/more natural under one measure than another, so having the transformation between the various measures comes in handy.
quantpie thank you!
Hi, 16:30 D(T) - D(t) is decomposed into two terms, and you mentioned the first expectation is 0, but the second is not. Does that means if the expectation of D(T)-D(t) is zero then this process is markovian ? Why 0 expectation can imply markovian property ? Thanks!
and when we say the vol sigma_f(s,T) is a deterministic function, is it because we are observing at T so the path is already fixed ?
No, if the expectation is zero it does not mean the process is Markovian (only that it is a Martingale) because we are only considering the first moment of the conditional distribution. However, since the first moment of D_T - D_t depends on the realisations of the Brownian up to and including time t, which are F_t-measurable, this is sufficient to conclude that the process is non-Markovian.
@@hihihi82 sigma_f(s,T) is deterministic, as opposed stochastic or constant, because it depends only on time. We are not observing at T - the equations are conditional on the filtration at time t < T. Deterministic does not mean measurable / observable, it just that we can perfectly predict what the value of the function will be because it depends only on time.
This saves my life! 5:56，the drift term, is that because of the leibniz integral rule?
yes implicitly does boil down to Leiniz integral rule. We went in reverse order though - from derivative to integral. The two sub components are coming from the left hand side of the blue box - equations at the top and bottom. thanks!
I love you
13:23