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2

Two volatility processes yield a higher flexibility of the model. This is of greater importance if one tries to price derivatives with different maturities in one single model. A additional volatility component helps to capture the term structure of volatility, which can depend greatly on time to maturity. See for example the VIX term structure from CBOE: ...


2

I think,the additional volatility factor,$v_2(t)$, provides more flexibility in modeling the volatility surface.We know $\rho$ controls the slope of the implied volatility.In the single-factor Heston model, $\rho$ is constant over maturities,In deed $$Corr[{dS}/{S\,,\,dv]}\;=\rho \,$$ which means that model has trouble providing an adequate fit to market ...


2

Loosely speaking, it can be seen as inserting an additional degree of freedom in the underlying's dynamics. This can be useful from a static perspective: with an additional lever to play on, one can hope to better capture the short term implied volatility smile, which "naive" stochastic volatility models (single volatility factor, no jumps) are known to be ...


1

Generally, the Wishart stochastic volatility model identifies the volatility of the asset as the trace of a Wishart process. Contrary to a classic multifactor Heston model, this model allows to add degrees of freedom with regard to the stochastic correlation. Thanks to its flexibility, this model enables a better fit of market data than the Heston model. ...


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 ...


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 ...


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*} ...


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 ...


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


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

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 ...


0

I think (1) is the issue. You need to compare market normal vols to normal vols implied by the sabr model. (2) is not the issue - these vols look reasonable. By the way we express normal vols in bp per annum, not percent!



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