# Tag Info

It is a probability measure in the mathematical definition of the term: returns results in the unit interval (price 0 for a claim that pays in an impossible event, and price $P\left(t, T\right) = \mathbb{E}^\mathbb{Q}\left(1 e^{-\int_t^T{r_sds}} \right)$ for a risk-free bond) and countable additivity property: if a claim pays either in event $A$ or in event $... 4 Originally "risk-neutral" is a term from economics describing the attitude of investors towards risk: if they are risk-neutral they only factor in the expected value of a decision and not the level of risk. "Risk-averse" would mean that they prefer investments with lower associated risk ceteris paribus (i.e. all else being equal). In quant finance risk-... 3 Shreve is a bit naughty here but, of course, he is right. When you have the risk-neutral measure$\mathbb{Q}$or$\tilde{\mathbb{P}}$, you can price derivatives as discounted expectation by the very definition of the risk-neutral measure (better called: equivalent martingale measure). So indeed, once you have$\tilde{\mathbb{P}}$, you can price derivatives ... 0 A priori, if those two$W_j$notations are in the same paragraph, they identify the same Wiener processes, under the same probability measure. So yes, they are seen as the different risk factors in your financial market. If you assume that your$W_j$'s are independent of one another, you can introduce correlation between your two variables (stock price and ... 2 For Risk-neutral Pricing to “work”, you need assumptions where risk elimination by trading financial instruments is possible : no counterparty risk, no transaction costs, continuous trading, continuous asset paths. If such assumptions are not fulfilled (which is the case in real markets ; however, for large banks they are sufficiently near from reality), ... 3 A call struck at$100$costs$2.97$, therefore a call with a strike higher than$100$must cost less than$2.97$. 5 Just to add to the answer by @KeSchn : There are at least two things going on here. First of all let$\{Q_i \}$denote a set of equivalent probability measures, which includes your$P$and$Q$above. Any$F^i(t)$defined as$F^i(t) = E_t^{Q_i} [P_T]$will be a martingale by application of the tower law. With the definition above, it will not be the case ... 5 You probably wonder whether$\mathbb{E}^\mathbb{P}[P_T\mid\mathcal{F}_t]= \mathbb{E}^\mathbb{Q}[P_T\mid\mathcal{F}_t]$. Note the$T$as index, i.e. the future unknown payoff and not the current price$P_t$. Now, why should$P_t$be a martingale under both,$\mathbb{P}$and$\mathbb{Q}$? Most likely, it is not. Indeed, the reason why you use$\mathbb{Q}$in ... 0 The current LIBORs (say O/N, 1wk, 1m, 2m, 3m, 6m, and 1y) refer to the borrowing cost for borrowing starting today (technically this is the spot date, which differ by currency,e.g. T+1, but we can call it today!). Each of these rates will have its own discounting curve. Let's focus on the 3 months rate. The current 3 month LIBOR refers to borrowing today ... 2 Let us start from your last equation, and focus specifically on the expectation. Assuming that the end date of each period is the start period of the next, the idea is to simplify it using conditional expectations. Since$t < t_{n-2}$, we can write using the tower property of conditional expectations:$\$ \begin{aligned} \Bbb{E}_{t}^{Q^{t_n}} \left[\prod_{...