# Tag Info

9

You don’t just need self-financing in a risk-neutral world but it’s a much more fundamental principle. If you look at a portfolio that is not self-financing, i.e. you can inject or withdrawal funds at any time, you can hedge any derivative easily. If you can always add the amount of money you need, then hedging becomes trivial. Thus, one requires the self-...

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Your mistake is actually made at the beginning: "Introducing a new process: $d\tilde{W}_t = dW_t +\frac{\mu-r}{\sigma} dt$" This is incorrect. Rather, $d\tilde{W}_t = dW_t -\frac{\mu-r}{\sigma} dt$ Otherwise, your derivation is correct. After correcting for the sign error, your final equation becomes $\Phi(x)=e^{-\lambda x-\frac{1}{2}\lambda^2 t}$. ...

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For the buyer of a forward contract the payoff is $S_T - K$ at time $T$ since at this date he pays $K$ and gets the underlying in exchange. Consider the following strategy: buy the stock $S$ and sell $K$ zero-coupon bonds with maturity $T$. At any time $t$, your portfolio's value is $$\Pi_t = S_t - KB(t,T)$$ In particular at time $T$, it is $S_T - KB(T,... 4 I suspect your expression of Gisanov has the wrong sign and should rather read: $$\tilde{W}_t = W_t + \frac{\mu-r}{\sigma} t$$ Equivalently, in differential form $$d\tilde{W}_t = dW_t + \frac{\mu-r}{\sigma} dt \tag{1}$$ Such that the dynamics under the physical measure $$\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$$ becomes, under the risk-neutral ... 4 The first fundamental theorem of asset pricing states (basically) a market is free of arbitrage if and only if there exists at least one equivalent martingale measure (EMM) The second fundamental theorem of asset pricing states (basically) that if a market is free of arbitrage and complete, then the equivalent martingale measure is unique. A market is ... 4 In practice, the self-financing condition can be regarded as an economic consequence of market competition. Take the perspective of an investment bank trading in hedgeable derivatives. If the hedging strategy is not self-financing, then it must be either: Generating cash outflows for the bank. It is therefore uneconomical for the bank to trade this product; ... 3 A non academic answer: In the real world, when dealers or professional counterparties trade options with each other, the option premium is not funded by the dealer's unsecured borrowing. Rather, options and other derivatives are usually subject to Collateral Agreements whereby 'safe' collateral is posted to cover exposure between counterparties. If cash ... 3 while in the normal world, we don't have to take the expectation of future cashflow Grossly incorrect. The formula or approach are exactly the same: the value of an asset is the expectation of its discounted cash flow. The only difference is the discount rate. If you have a nonlinear instrument such as a vanilla Euro call option, it's not clear what should ... 3 As with everything else it is determined by competition: little or no competition => very high fees (or more correctly large bid-ask spreads). That is one reason why many IB try to develop new derivatives: they can be very profitable when no one else trades them yet. Then the cost comes down somewhat when competitors come in. Lack of transparency in pricing ... 3 Your question is not clear. What you might want to say is what distribution should the futures price follow, under the risk-neutral or physical probability measure. In this sense, it will depend on your intention. For potential future exposure, you may want to use the physical measure for the price evolution, while the distribution will depend on your model ... 2 First lets analyse the claim that$\frac{(S_t - F(t,T)}{S_{0}}$is a martingale under a given risk neutral measure$P^{*}$. Recall that the crucial property of a martingale is that at some point in time$t$, a process$\tilde{S}_{t}$is a martingale iff for some time$t+\Delta$, the expected value of$\tilde{S}_{t+\Delta}$is$S_t$. So lets start, assume we ... 2 After a lot of guess work, I think can try and answer what I think might be your question. First, note that at maturity the forward equals the spot:$F_T^T = S_T$so I am not sure what you mean by "forward price strike". I think you mean that your have forward prices of calls and puts. If you chose a model for your index$S$and the rates, then the ... 2 It really simplifies your life when dealing with valuation. As stated already a non-self-financing portfolio either generate or absorb cashflow. Such cash flows would need to be taken into account when valuing a certain derivative based on replication. So, in general, is way easier to just deal with a self-financing portfolio. On the other hand, when you ... 1 A brief educational note and then where you can find the info... As a first step, set the expected payoff equal to 0 where prob_D = probability of default, cur_Px = current price, mat_Px = maturity payment, and R = recovery. Therefore prob_D * (recovery - cur_Px) + (1 - prob_D) * (mat_Px - cur_Px) = 0 results in prob_D = (cur_Px - mat_Px) / (R - ... 1 In the first$\Delta \Pi$expression we are trying to eliminate risk from the portfolio. It just so happens that to do the delta hedging in this case we need to take into account the dividend. The second$\Delta \Pi$expression comes from the no-arbitrage principle, which is the same (as is its equation$\Delta \Pi = r \cdot \Pi \cdot \Delta t$) be it ... 1 You could for example parametrize your risk-neutral density$\hat{g}(S_T=x)$as a polynomial: $$\hat{g}(S_t=x)=\sum_ia_ix^i$$ and solve the program for a chosen polynomial order$n =\max i: \begin{align} & \min_{(a_i)}\left[\sum_k\left(\sum_i a_ix_k^i-\hat{g}(S_t=x_k)\right)^2\right] \\[3pt] & \ \forall \ k, \ \sum_ia_ix_k^i\geq0 \end{align} ... 1 Perhaps other memeber of qSE are going to correct me, but I think the following rule of thumb is useful. Whenever you have a doubt, try to forget that a pricing measure is a probability measure. This is just a pricing tool: originally for any option/derivative/contingent claim we'd like to know its price, so we introduce a map\pi:X\to \Bbb R$such that$\...

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Five years late to the party, but let me put my two cents in for intuition. Let $W^P_t \equiv W_t$, and $W^Q_t \equiv \tilde{W_t}$ to easier remember under what measure either one is a standard Brownian motion. Also, let $\lambda := \frac{\mu-r}{\sigma}$, which will be used as abbreviation below. Then, $\frac{dS_t}{S_t} = \mu dt + \sigma dW^P_t$ $\... 1 I believe this is the way that Björk proposes, however I believe "my" way below is more elegant. The trick in Björk's case is to realize that in each "iteration" we get an expeted value:$\$V_0(0) = \frac{1}{1+R}\left(q_u V_{1}(1) + q_d V_1(0) \right) \\ = \frac{1}{(1+R)^{2}} \left( q_u^2 V_2(2) + 2q_uq_d V_2(1) + q_d V_2(0) \right) \\ = \frac{1}{(1+R)^{2}}E^...

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