Bounds on option

Hello everyone, I would really appreciate some help about bounds on option… i cant understand it by mylself for days now… 1. If the max value of the euro put must be Xe^-rt, because the value will only be X at expiry for the writter, so why the max value of the call is not also discounted ? max call value = Se^-rt would be logical because the amercan call would be S (higher) tough. 2. is there a difference between minimum value and lower bound value ? because sometime it is told that the value of an option is zero… BUT the lower bound value may be more. It is confusing, so many arbitrage possibilities in every ways… Thanks by advance, Coritani

Well it kind of is in a way, the present value of the stock is todays price.

So as opposed to the put proceeds, where you invest at the risk-free rate, the call porceeds are invested in the underlying.

So if c>S(0), then you could sell the call today and buy the underlying.

Then you have made c-s(0)>0, upfront.

If the call expires in the money or out of the money at expiration, your profit is still positive.

Hope that helps…

And minimum value and lower bound are the same thing.

Thanks for trying to help me. I know that the discounted value of the stock price today is the stock price today, my concern is only about the call value/premium/price But if there is a stock trading at 50 today and a call whose strike is 20, then the intrinsic price of the call would be 30. BUT 30 is the value of the call only at maturity so it should be sold at 30e^-rt… AND if the stock trades at 50, and that the strike is zero, then the price of the call should be 50e-rt : that is to say the discounted price of the current price of the stock (sounds weird…), which would be the maximum possible

I think the reason is that the arbitrage opportunity only presents itself at the value of S(0), otherwise you will not have enough proceeds from short sale of the call option to even purchase 1 of the underlying to cover your exposure.

The whole point of arbitrage is to gurantee a risk-free profit with no negative net cash flow at any point. So if the price was not at least S(0), there would be a negative cash flow at inception.

Also, intrinsic value is different from maximum value.

If it trades at intrinsic value, there is defintely no arbitrage opportunity (unless intrincisc value equals maximum value).

Tha maximum value is only there to say it CANNOT trade above that, but it definitely does not mean it will trade at that price or anywhere near it.

I think i begin to understand this point… i need to think about it a little bit more… But thinking about it, i realized that i got to a point further in the curriculum : the lower bound. My thought is as following : If i own the stock (So = $50), i agree to sell it in 1 year at 50 if, at least, the call bring me the RFR during this period right ? (the cost of carry the stock i guess) So if the price is 50 and strike is 50, the intrinsic price of the written call is not zero but the RFR computed on my stock price… Here is what i dont understand : I want the RFR on my stock price, not on the strike, because if the strike is 30, then i get the RFR on only 30… the formula is “call = max(0, So - Xe^-rt)…” Shoudnt the stock holder want to get the RFR on the stock price ? As the strike goes to zero, according to the formula, the RFR benefit tend to decrease until zero too. But the day the call is sold, the price of the stock is still 50, and the stockholder should want the RFR on this base…

i understood finally : he doesnt need to get the RFR on 50 because he is given 50 - 30 = 20 So he reinvest 20 -> 20*(1+RFR)^t & he got from the call the RFR on 30 only the total is that he gets the RFR on 50