Hot answers tagged

7

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process) $$ df(t,X_t) = ...


6

[Short answer] No closed-form formula in general. You need to resort to numerical methods. Monte Carlo is preferred by most practitioners but you could also use Finite Difference schemes (and sometimes even Fourier inversion techniques depending on the model used and the instruments to be priced). [Long answer] One usually distinguishes between 2 classes ...


3

Your problem probably comes from the notations used. Let the Moment Generating Function (MGF) of a random variable $X$ be defined as $$ M_X(u) := E[e^{uX}] $$ From this definition, it entails that $$ E(X^n) = M_X^{(n)}(u=0) = \frac{d^{n} M_X}{ d u^{n}}(u=0) $$ Knowing this, the function $$ f_{\lambda}(t,r)=E[e^{-\lambda {r_{T}}}|r_t=r] $$ can be ...


3

Here's my 2 cents: a) Conditional expectations can always be seen as martingales (this is a direct consequence of the tower property). Thus, we here have that $$ M_t := E^*[e^{-\lambda {r_{T}}}|r_t] $$ is a martingale. Applying Itô's lemma to $M_t = f_{\lambda}(t,r_t)$ as you did is a good starting point. But doing this, leaves you with an SDE, not a ...


3

Let $$ f_{\lambda}(t,r)=E^{(t,r)}\left[e^{-\lambda r_{T}}\right] $$ where $E^{(t,r)}$ denotes the expectation conditional on $r_{t}=r$. We assume $f$ is smooth for the remainder. Let $\theta=T\wedge\inf\left\{ s>t\colon\left|r_{s}-r\right|>1\right\} $. By the Markov property of $\{r_{t}\}$, $$ ...


2

The answer is yes. In fact, there always exist a 'Black Scholes like' formula. Easy to show too. If the risk neutral distribution of the price has cumulative density $P$ and probability density $p$, then $$ E(S-K)^+=E((S-K)\ 1_{S>K})=E(S\ 1_{S>K})-K\ E(1_{S>K}) $$ The second expectation is just $P(K)$, ie the probability that the option ends up in ...


2

From Equation (6), $B(t,T)=-t+c(T)$ for some function $c(T)$. $1=P(t,t)=e^{-A(t,t)-(c(t)-t)r_t}$ or $A(t,t)+(c(t)-t)r_t=0,\,\forall (r_t,t)$. So $c(t)=t, A(t,t)=0,\forall t$. For Equation (8) you have missed the square on $\sigma$ and a factor of $\frac13$. Then you just need to substitute in the function for $b(s)$ and integrate the following to get the ...


2

For starters, the short rate model you mention in equation (1) is Cox-Ingersoll-Ross while the bond price in equations (2)-(4) correspond to the Vacisek model. So there is a problem somewhere, I would go for a typo in (1). Second, what you wrote seems fine to me, so there must definitely be yet another typo in your solution manual. Note that if there is no ...


2

You almost get there. However, you ca not conclude that $\rho^2$ is a constant based on $(10)$. Note that, from your $(7)$ and $(8)$, \begin{align*} \frac{\rho(z_t)^2}{\beta} e^{\beta \tau} (e^{\beta \tau} - 1) = -h'(\tau)+e^{\beta \tau}h'(0). \end{align*} Taking derivative with respect to $\tau$ on both sides, we obtain that \begin{align*} ...


1

Local vol model gives a "too shallow" forward skew. Derivatives of which the price are depending on the forward skew will be mispriced. If i remember correctly, Hagan's paper



Only top voted, non community-wiki answers of a minimum length are eligible