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Option on risky bond
Posted: February 2nd, 2012, 2:04 am
by shunvwu
Hi Guys,When using the Black's(76) Model in pricing the bond option price, the formula is defined as:c=P(0,T)[Fo*N(d1)-K*N(d2)], in which Fo is the forward price of the bond. Supposing the risk-less rate is r1 while the yield of bond is r2, in my understanding Fo should be S*exp(r2*T). Is it right? And how about P(0,T), should it be exp(-r1*T) or exp(-r2*T)? I tried to search papers but didn't find anyone discussing this issue detailedly. I'm quite confused about this issue, could anybody kindly give me some advices, or are there any papers talking about this issue?Thanks very much!
Option on risky bond
Posted: February 2nd, 2012, 7:27 am
by daveangel
no - the forward price of an asset is determined by the "risk-free" rate.
Option on risky bond
Posted: February 2nd, 2012, 8:25 am
by shunvwu
Hi daveangel,Thanks for your reply. Do you mean that the Fo should be S*exp(r1*T), and P(0,T) equals to exp(-r1*T)? Thus the bond option price has nothing to do with r2?
Option on risky bond
Posted: February 2nd, 2012, 10:40 am
by daveangel
the t-time forward price of the bond will be roughly S * exp((r1 - r2)*t) where S is the spot price, r1 is the "risk-free" rate and r2 is the yield on the bond
Option on risky bond
Posted: February 2nd, 2012, 12:17 pm
by shunvwu
Hi daveangel,I'm sorry but I can't understand the formula of forward price. When I mentioned the "yield" r2, I meant that this bond is risky. Because the bond has default possibility, its price is discounted by r2 from par to S on today instead of using r1 for discounting(supposing the bond is zero-coupon bond). Thus I thought the forward price of this bond should be S*exp(r2*T). And when the bond defaults, the option price is zero, thus it seems the option price is also risky and P(0,T) should be exp(-r2*T).But in risk-neutral world, it indeed consider an asset's yield is the risk-free rate r1, thus from this point, the forward price as well as the option price should have nothing to do with r2, so the Fo=S*exp(r1*T) and P(0,T)=exp(-r1*T). I think both of the above two are reasonable thus I'm quite confusing. But I really can't understand why Fo equals to S * exp((r1 - r2)*t) in this situation In my opinion, S * exp((r1 - r2)*t) happens in such situation like a stock option when the stock has dividend yield. Could you please kindly explain more?
Option on risky bond
Posted: February 2nd, 2012, 12:28 pm
by daveangel
I would think that the price of jump to default would be in the implied volatility...