How d2 in Black-Scholes becomes PD in Merton model
Uploader Comments (bionicturtledotcom)
All Comments (14)
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I was thinking about the relationship between time and distance to default. It looks like the bigger the maturity, the bigger the distance to default (ceteris paribus). Does it make sense to assume that? However, this works only if the second term in the numerator is positive, so the price grows more than half of the variance.
Am I correct?
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amazing thank you !!! life saver as usual
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and really irritated with a few comments over here..guys..we are getting such valuable education in youtube for free (thanks Mr.Harper).. we cannot be complaining
Mr.Harper's videos are supplements; we should not expect him to teach everything over here.
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wow...what a clear explanation.. Mr.Harper, you are awesome.
I am learning finance on my own; have watched Paul Wilmott's lectures in you tube, Horward lectures.. but nothing is as clear, concise and understandable as your short 10 min videos.. thanks a lot
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wow amazing clarity on d2. i pondered through my notes without much clarity but this video just connected the dots. thanks for the great vid!
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what is blackshole model?you should have explained it first
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i wish i can find a video explaining d1 as well ... this video was really good ... cheers
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thank you, really helpful
at 4:00 you say the firm would have to drop by 26% before default. 0.74*130 = < 100 though, so I don't see why it should not bee a lot less than 26%
KoalaBearWarrior 1 week ago
@KoalaBearWarrior Like Black Scholes, the rates are all expressed in continuous frequency. At T0, LN(130/100) ~ 26%. At T1, LN(134/100) ~= 29.2. To use 0.74*130, is to use discrete: 134/100 - 1 = 34%. FWIW, they reconcile with R(c) = k*LN(1+R(k)/k). In this case, case R(c) = 1*LN(1 + 34%) = 29.2%
bionicturtledotcom 1 week ago