# Option question

An analyst is evaluating a European call option with a strike price of 25 and 219 days to expiration. The underlying stock is currently trading for \$29, and the analyst thinks that by the option expiration date the stock will be valued at \$35. If the risk-free rate is 4.0%. what is the lower bound on the value of this option? A \$0 B \$4.00 C \$4.58

C

Answer is C, but in true Schweser style you have to assume quite a few things (risk neutral world). I actually thought it was A.

you dont have to assume anything here, Option value = Intrinsic value + Time value, Intrinsic value (call) = max ((spot - PV(strike)), 0), even if time value were 0 (zero vol), the intrinsic value is definitely above 0, and in fact above 4 because of tmv. I saw L2 forum, and based on the threads, options are quite heavy there (delta hedging, greeks, gamma limits, and so on) I recommend to understand this… My job is in derivatives, so easy for me …

I’m confused…I thought it was A too? I agree with pfcfaataf that there certainly is intrinsic value here, but whats to stop the option ending up out of the money and therefore being bounded by zero? If the question was what is the fair value, or intrinsic value I’d agree with the choices. But I don’t get this.

kurupt, options are always valued more than zero or otherwise they can’t be sold. But the pay off’s may be zero or negative all the time. Unless the markets see intrinsic value or time value, options cannot be priced. If they are not readily in the money, they have atleast time value – hope that the price of the underlying rises before the option expires. hth

ov25 you’ve just showed that 0 is the lower bound of an option price. What I actually thought the question was asking for was the lower bound of the options value at expiration (which is obviously 0). I’m going to have to go over the option chapter i think because I actually skipped it assuming knowledge of binomial pricing, black scholes pricing and martingale pricing would be more than enough to get through it - obviously I need to brush up on the basics!

of course, minimum value of the option anytime (any spot, any rates, any vol) during the life can theoretically go to zero. the value of the option at given time, given spot, given rate has lower bound equal to … you know…

This reminds me of the time during my degree someone asked me to help them with some stats (half my degree was stats) and I couldn’t actually do it because it was so simple that I had completely forgotten.

I stand corrected… I got confused between ‘lower bound’ and percieved value of an option. Lower bound = Max(0, underlying-PV of strike rate@rfr) In this case, it is in the money so the lower bound CANNOT be zero.

can somebody show me how to get 4.58 in terms of calculation? Thanks

I’m wondering the same thing. When I took \$4 at the 4% over 219 days, I only got \$4.10. Confused.

Its a European option: 29 - (25/(1.04^(219/365) = 4.58 UGhhh Shortcut: European lower bound must be = or > the American lower bound. If this were american, it would already be worth \$4 plus the time value portion, so must be > \$4. the european must at least equal the american, so the number >\$4 is the only correct choice.

Min European = max (0, S- (X/((1+rf)^.6))) 219/365=.6 X=Strike Price S=Stock Price This gets you 4.58

june, please excuse my ignorance. Would my calculation of \$4.10 have been correct for an American option?

here you go billy Int the calc i/y =4/365 n=219 fv=25 PV= you get 24.41 29-24.41= 4.59

i dont think so…schweser is saying the lower bound formulas are the same for ameican and european calls. european lower limit must be = or > American.

pistan669 Wrote: ------------------------------------------------------- > here you go billy > > > Int the calc > > i/y =4/365 > > n=219 > > fv=25 > > PV= you get 24.41 > > 29-24.41= 4.59 Ohhhhh i like this method. Will it always yield the correct lower limit? NICE!!