# Stock forward price argument

Hi I am strangling to understand where is the mistake with the following strategy. Can anyone help me with the following argument?

Assuming a stock price follows geometric Brownian motion then the expected value, under the physical measure, is $S\exp(μt)$. If the forward price is $S\exp(rt)$ then if I always long the forward I will be profitable in the long run ( i understand there is a risk envolve and that this difference it can be explain by the market price of risk. I also understand that if the price of the forward is $S\exp(μt)$ there will be arbitrage by borrowing money buying the stock and selling it at the forward price.) But still under those assumptions if the price of forward is $S\exp(rt)$ by entering long I will be profitable in the long run.

Can anyone point out where my mistake is?

• Do you want to know what is wrong with your idea, or if there is something wrong with your idea given the constraints you have imposed upon the problem. – Dave Harris Jul 12 '18 at 16:38
• You need to clarify your question, what are you trying to achieve precisely? Anyway, to help clarify things for you, the price of the forward is given by a no arbitrage argument. Buying the stock and holding it until $t$ while selling the forward is not an arbitrage, you have a non-zero probability to lose money. – byouness Jul 13 '18 at 11:35