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[thm]{Exercise}
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\begin{document}
\title{ABC Trial}
\maketitle
\section{P1}
\[
dS_t = rS_t dt + \sigma S_t dW_t
\]
Note: This is a dividend paying stock, but I was not given any information about the dividend for this particular problem. So let's proceed as follows:
\begin{align*}
PV_t
& =E^Q[e^{-r(T-t)}\frac{1}{S_T}]\\
& =e^{ - r(T - t)} E^Q [(S_t e^{(r - \frac{1}{2}\sigma ^2 )(T - t) + \sigma W_{T - t} } )^{ - 1} ]\\
& =e^{ - r(T - t)} S_t ^{ - 1} E^Q [e^{ - (r - \frac{1}{2}\sigma ^2 )(T - t) - \sigma W_{T - t} } ]\\
& =e^{ - r(T - t)} S_t ^{ - 1} e^{ - (r - \frac{1}{2}\sigma ^2 )(T - t) + \frac{1}{2}\sigma ^2 (T - t)}\\
& =e^{-2r(T-t)+\sigma^2(T-t)}S_t^{-1}.
\end{align*}
a) This contract has risk exposures to r, to $\sigma$, and to $S_t$. Therefore, ideally speaking the hedging should consider all the risk exposures.Let's take Delta-hedging as an example:
\[
\frac{{\partial PV_t }}{{\partial S_t }} = e^{ - 2r(T - t) + \sigma ^2 (T - t)} \frac{{ - 1}}{{S_t^2 }} \buildrel \Delta \over = \Delta
\]
As you can see, $\Delta$ depends on S and t and other parameters, which means the hedge will need to change continuously.
Also, the continues hedging incurs huge transaction costs.
Also, the asset price may not follow log-normal distribution at all.
b)$\Gamma\triangleq\frac{\partial ^2PV_t}{\partial S_t^2}=e^{-2r(T-t)+\sigma^2(T-t)}\frac{2}{S_t^3}>0$
So there is convexity. So it could be a long/short Gamma trade.
\section{P2}
\begin{align*}
max(C_1,C_2)
& =max((S_{1T}-K_1)^+,(S_{2t}-K_2)^+)\\
& =max(S{1T}-K_1,S_{2T}-K_2,0)
\end{align*}
Called dual-strike Rainbow option.
a)$
\left\{ \begin{array}{l}
F_t + rS_1 F_1 + rs_2 F_2 + \frac{1}{2}S_1^2 \sigma _1^2 F_{11} + \frac{1}{2}S_2^2 \sigma _2^2 F_{22} - \rho \sigma _1 \sigma _2 S_1 S_2 F_{12} - rF = 0 \\
F(T,S_{1T} ,S_{2T} ) = \max (S_{1T} - K_1 ,S_{2T} - K_2 ,0) \\
\end{array} \right.
$
b) There is no close-form pricing formuke for dual-strik Rainbow option.
Of course, if $K_1 \equiv K_2 = K$, then the payoff
\begin{align*}
F(T,S_{1T},S_{2T})
& =max(S_{1T}-K,S_{2T}-K,0)\\
& =max(S_{1T},S_{2T},K)-K.
\end{align*}
Then it becomes a Rainbow option of "Best-of-plus-cash" type minus a bond position.
Stulz(1982) Studied this type of option and found close form solution.
Let
\begin{align*}
\sigma_A& =\sqrt{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}\\
\rho_1 & =\frac{\rho\sigma_2-\sigma_1}{\sigma_A}\\
\rho_2 & =\frac{\rho\sigma_1-\sigma_2}{\sigma_A}.
\end{align*}
c)
1) Fortunately her we have non-time dependent volotilities, otherwise we will have non-combining trees.
Because we have two underlying random sources, we need to use 2D Brownian Motion. Let it be $W_t=(W_t^1,W_t^2)$, we could generate
\[
\left\{ \begin{array}{l}
Z_t^1 = W_t^1 \\
Z_t^2 = \rho W_t^1 + \sqrt {1 - \rho ^2 } W_t^2 \\
\end{array} \right.
\]
and then generate
\[
\left\{ \begin{array}{l}
S_1 (t) = S_1 (0)\cdot e^{r - \frac{1}{2}\sigma ^2 )t + \sigma \sqrt t Z_t^1 } \\
S_2 (t) = S_2 (0)\cdot e^{(r - \frac{1}{2}\sigma ^2 )t + \sigma \sqrt t Z_t^2 } \\
\end{array} \right.
\]
Let time step be $\delta t$, for the first time step, we have four possible states:
\[
(\sqrt {\Delta t} ,\sqrt {\Delta t} ),(\sqrt {\Delta t} , - \sqrt {\Delta t} ),( - \sqrt {\Delta t} ,\sqrt {\Delta t} ),( - \sqrt {\Delta t} , - \sqrt {\Delta t} )
\]
For jth step we have $j^2$ states.
We have only three hedging instruments while at each stage we have four possible states. So that's imcomplete market. But it doesn't matter because we only use tree to approximate the already arbitrage-free rick neutral measure.
Of course, we could be smarter and only use three steps at each state:
\[
\sqrt 2 \Delta t(0,1),\sqrt 2 \Delta t(\frac{{\sqrt 3 }}{2}, - \frac{1}{2}),\sqrt 2 \Delta t( - \frac{{\sqrt 3 }}{2}, - \frac{1}{2})
\]
This tree is a recombining tree.
(2): if we still use tree then there will be an explosine number of nodes. So we will have to use Monte Carlo. Of course, in using Monte Carlo, we haave to handle conelased state variables appropriately. Variance reduction and low-discrepancy sequences are frequently preferred alternately.
\section{P3}
(a)
\begin{align*}
S_T & = S_0 e^{(r - \frac{1}{2}\sigma ^2 )T + \sigma W_T } \\
Q(S_T \le x) & = Q(S_0 e^{(r - \frac{1}{2}\sigma ^2 )T + \sigma W_T } \le x) \\
& = Q((r - \frac{1}{2}\sigma ^2 )T + \sigma W_T \le \log (\frac{x}{{S_0 }})) \\
& = Q(W_1 \le \frac{{\log (\frac{x}{{S_0 }}) - (r - \frac{1}{2}\sigma ^2 )T}}{{\sigma \sqrt T }}) \\
& = N(\frac{{\log (\frac{x}{{S_0 }}) - (r - \frac{1}{2}\sigma ^2 )T}}{{\sigma \sqrt T }}) \\
Q(S_T \in dx) & = N'(\frac{{\log (\frac{x}{{S_0 }}) - (r - \frac{1}{2}\sigma ^2 )T}}{{\sigma \sqrt T }})\cdot\frac{1}{{\sigma \sqrt T }}\cdot\frac{1}{x}.
\end{align*}
(b)
\begin{align*}
Q(S_T \in dx) & = \frac{1}{{\sqrt {2\pi } }}e^{ - \frac{{(\log (\frac{x}{{S_0 }}) - rT + \frac{1}{2}\sigma ^2 T)^2 }}{{2\sigma ^2 T}}} \cdot\frac{1}{{\sigma \sqrt T }}\cdot\frac{1}{x} \\
& = \frac{1}{{\sqrt {2\pi } }}\exp ( - \frac{{(\log (\frac{x}{{S_0 }}) - rT)^2 }}{{2\sigma ^2 T}} - \frac{{(\log (\frac{x}{{S_0 }}) - rT)}}{2} - \frac{1}{8}\sigma ^2 T)\cdot\frac{1}{{\sigma \sqrt T }}\cdot\frac{1}{x} .
\end{align*}
For find x, the PDF looks like a skewed bell-shaped curve with a peak.
For fixed x, and if we let $\sigma$ vary,
then the exponential part has a peak at $
\sigma ^2 T = \sqrt {4(\log (\frac{x}{{S_0 }}) - rT)^2 } = 2\left| {\log (\frac{x}{{S_0 }}) - rT} \right|
$
And the exponential part is gymmetric to the left and to the right side of this this $\sigma^*$ point.
Together with the $\frac{1}{\sigma\sqrt{T}}$ term, it can be seen that $\frac{exp(\cdots)}{\sigma \sqrt{T}}$ decreases is $\sigma$, for fixed x.
But $
\sigma (k) = \sigma _0 + \varepsilon \log (\frac{{S_0 }}{k})
$, which is decreases in K, therefore the pdf increases in k, for fixed x.
c) The shape of the vol-skew is okay in the equity market. However everything is deterministic here, which doesn't reflect the other dimension of the vol-surface i.e. the time-to-maturity dimension.
The implied vol form in (b) is impossible in the market.
d) Yes. There are a few problems with B formulation:
\begin{description}
\item[\textcircled{1}] Lord vol is deterministic; not suitable for fornord start aptuns, where values depend on random vol itself.
\item[\textcircled{2}] It does not vary with time-to-maturity.
\end{description}
e)\begin{align*}
E^Q [e^{ - r(T - t)} I_{\{ S_T > K\}} ]
& = - \frac{\partial }{{\partial k}}E^Q [e^{ - r(T - t)} (S_T - k)^ + ] \\
& = - \frac{\partial }{{\partial k}}C(k) .
\end{align*}
Therefore we just have to take derivative of call price as function of K w.r.t K. We could do this because we are given call optionprices with set of strike form a continum.
We could do this by replicating the payoff of digital using liquidly traded call options.
\section{P4}
a)\[dS_t=S_t(r-q)dt+\sigma S_tdW_t\]
where q is the continuons dividend yield.
b) \begin{align*}
S_T & = S_0 \prod\limits_{i = 1}^n {(1 - P(t_i ))\cdot} \exp (\sigma ^{w_T + (r - \frac{1}{2}\sigma ^2 )T} ) \\
S_T & = (S_0 - \sum\limits_{i = 1}^n {e^{ - rt_i } c(t_i ))} \cdot\exp (\sigma W_T + (r - \frac{1}{2}\sigma ^2 )T) \\
\end{align*}
c)The price of the call is $
e^{ - rT} (S_0 e^{(r - q)T}N(d_1 ) - kN(d_2 ) ) \triangleq $ let's define this to be F,
where \[
\begin{array}{l}
d_1 = \frac{{\ln (\frac{F}{K}) + \frac{{\sigma ^2 }}{2}T}}{{\sigma \sqrt T }}, \\
d_2 = \frac{{\ln (\frac{F}{k}) - \frac{{\sigma ^2 }}{2}T}}{{\sigma \sqrt T }} \\
\end{array}
\]
For the other schemes, we just have to change F into the form of $
S_0 \prod\limits_{i = 1}^n {(1 - \rho (t))\cdot e^{rT} }
$
and $
(S_0 - \sum\limits_{i = 1}^n {e^{ - rt_i } c(t_i ))\cdot e^{rT} }
$
For comparison, we just have to compare.
\begin{center}
$e^{-qT}$ vs. $\prod\limits_{i=1}^n{(1-P(t_i))}$ vs.
$
1 - \frac{{\sum\limits_{i = 1}^n {e^{ - rt_i } C(t_i )} }}{{S_0 }}
$
\end{center}
d) same BS formulae.
We just have to take the root mean of integrated time average of $\sigma$, i.e.
\[
\sigma _{0,i}^2 = \frac{1}{{t_i - t_{i - 1} }}\int_{t_{i - 1} }^{t_i } {\sigma _S^2 ds}
\]
Therefore, we could map the "regime-switching" time dependent volatility into piece-wise constant volatility. We could still use the BS formulae.
\section{P5}
a) A factory is a method, an object, or anything else that's used to instantiate other objects.
b) Let's say we are writing a vanilla option pricer with given payoff, specified by its name in string using traditional method, we could put string names of the payoffs into header file and use a switch statem clearly, every time we add a new pay-off, we have to modify the header and he switch statement.
We could use "object factory" to solve this problem. For each pay-off class, we write an auxiliary class whose constructor registers the pay-off class with our factory.
Therefore, we could add new payoffs without changing any existing files. We just have to simply add new files to the project.
\section{P6}
a)If the Bamer is monitored continuously, then it's relatively easy to numerical PPE approach, since we just have to apply boundary conditions at the barrier.
For barrier that is monitored discretely, Monte Carlo simulation is unbiosed (provided that sampling dates for our simmulations coincide with barrier observation times), but slow to converge.
There are serveral papers used implicit or Crank-Nicolson method to hondle Bariner options.
b)For a convertible bond that offers early exercise, it has been the retically shown that the early exercise trigger is the first hitting time of a barrier and this barrier can be time-dependent. Thus the similarity between the two models is theorectically justified.
\section{P7}
\begin{align*}
PV_t
&= E^Q [e^{ - r(T - t)} (\log S_T )^2 ] \\
& = E^Q [e^{ - r(T - t)} (\log S_t e^{(r - \frac{1}{2}\sigma ^2 )(T - t) + \sigma W_{T - t} } )^2 ] \\
& = e^{ - r(T - t)} E^Q [(\log S_t + (r - \frac{1}{2}\sigma ^2 )(T - t) + \sigma W_{T - t} )^2 ] \\
& = e^{ - r(T - t)} E^Q [(\log S_t + (r - \frac{1}{2}\sigma ^2 )(T - t))^2 + 2(\log S_t + (r - \frac{1}{2}\sigma ^2 )(T - t))\sigma W_{T - t} + \sigma ^2 W_{T - t} ^2 ] \\
& = e^{ - r(T - t)} \{ (\log S_t + (r - \frac{1}{2}\sigma ^2 )(T - t))^2 + \sigma ^2 (T - t)\} \\
\end{align*}
a)The contract has risk exposures to r, $\sigma$, and $S_t$.
Here we use Delta hedging as an example:
\[
\frac{{\partial PV_t }}{{\partial S_t }} = e^{ - r(r - t)} 2(\log S_t + (r - \frac{1}{2}\sigma ^2 )(T - t)\frac{1}{{S_t }}
\]
Drawbacks:
\begin{description}
\item[\textcircled{1}] The hedge has to be continuous.
\item[\textcircled{2}] Continuous hedging incurs huge transaction costs.
\item[\textcircled{3}] The asset price may not follow log-normal distribution at all.
\end{description}
\section{P8}
a) First of all, we note that it's never optional to early exercise for a
non-dividend paying American call.
The decision about whether to early exercise or not is made based on
comparison of $C_t $ vs. $S_t - k$.
For dividend paying stock, away from the ex-dividend and expiration dates,
the American call option value strictly greater than the exercise value of
the option. That's to say, $C_t > S_t - k$ when away from ex-dividend and expiration dates.
Now if vol $ \uparrow $, $C_t \uparrow $, so you are less likely to early
exorcise the call.
\noindent
b) If $r \uparrow $, $C_t \uparrow $, because the strike price K you are
potentially giving up has a smaller present value. Therefore, you will give
up less, so $C_t \uparrow $.
Therefore you are less likely to early exercise the call.
\noindent
c)If dividends are higher, you are more likely to early exercise because you
want to capture the dividend.
\noindent
d)Now for a convertible bond:
When you don't exercise, you retain the bond cash flows plus the future
option to convert.
When you do exercise, you will lose the bond cash-flows plus future
conversion option but you get the stock plus its future dividend payments.
\begin{description}
\item[---] when vol$ \uparrow $, the option of future convertion is more valuable, so you are less libely to exercise.
\item[---] When interest rotes $ \uparrow $, the bond cash-flows gets discounted more,the future conversion option part increases in value, but the stock's future dividend paymants get discounted more too. So it's hard to judge whether in this case I'm more likely or less likely to exercise.
\item[---] When dividends $ \uparrow $, the future conversion option part drops in value, while the future dividend payments part increases in value.
\end{description}
Therefore I'm more likely to exercise.
\section{P9}
The problem:
Because the integration is w.r.t. the x, instead of the strike directly ,
therefore the weights in the ``result'' variable should be
$\frac{1}{(strike)}$ instead of $\frac{1}{(strike)^2}$.
Also,maybe it's better to use a vol-smile model.
Here is the derivation of the correct form:
\begin{align*}
\frac{ds_t }{s_t } & = \mu _t d_t + \sigma _t dz_t \\
d(\log s_t ) & = (\mu _t - \frac{\sigma _t ^2}{2})d_t + \sigma _t dz_t \\
\frac{ds_t }{s_t } - d(\log s_t ) & = \frac{\sigma _t ^2}{2}dt \\
\end{align*}
The total variance
\begin{align*}
& = \frac{1}{T}\int_0^T {\sigma _t ^2dt} \\
&= \frac{2}{T}\left\{ {\int_0^T {\frac{ds_t }{s_t } - d(\log s_t )} }
\right\} \\
&= \frac{2}{T}\left\{ {\int_0^T {\frac{ds_t }{s_t } - \frac{\log s_T }{\log
s_0 }} } \right\} \\
\end{align*}
Now $\mbox{ }\frac{\log s_T }{\log s_0 } = \log \frac{s_T }{\bar {s}} + \log
\frac{\bar {s}}{s_0 }$
Let's look at a twice differentiable payoff f(s):
\begin{align*}
f(s_T ) &= f(\bar s) + f'(\bar s)(s_T - \bar s) + \int_0^{\bar s} {f''(y)(y - } s_T )^ + dy + \int_{\bar s}^{ + \infty } {f''(y)(s_T - } y)^ + dy{\rm{ }} \\
&= f(\bar s) + f'(\bar s)\left[ {(s_T - \bar s)^ + - (\bar s - s_T )^ + } \right] + \int_0^{\bar s} {f''(y)(y - } s_T )^ + dy + \int_{\bar s}^{ + \infty } {f''(y)(s_T - } y)^ + dy.
\end{align*}
Therefore
\begin{align*}
\log (s_T ) &= \log (\bar s) + \frac{1}{{\bar s}}\left[ {(s_T - \bar s)^ + - (\bar s - s_T )^ + } \right] - \int_0^{\bar s} {\frac{1}{{y^2 }}(y - } s_T )^ + dy - \int_{\bar s}^{ + \infty } {\frac{1}{{y^2 }}(s_T - } y)^ + dy \\
\log (\frac{{s_T }}{{\bar s}}) &= \frac{1}{{\bar s}}(s_T - \bar s) - \left[ {\int_0^{\bar s} {\frac{1}{{y^2 }}(y - } s_T )^ + dy + \int_{\bar s}^{ + \infty } {\frac{1}{{y^2 }}(s_T - } y)^ + dy} \right] \\
E^Q \left[ {\int_0^T {\frac{{ds_t }}{{s_t }}} } \right] &= rT \\
E^Q \left[ {\frac{1}{T}\int_0^T {\sigma _t ^2 d_t } } \right] &= \frac{2}{T}\left\{ {rT - \log \frac{{s_T }}{{\bar s}} - \log \frac{{\bar s}}{{s_0 }}} \right\} \\
&= \frac{2}{T}\left\{ {rT - E^Q \left[ {\frac{{s_T }}{{\bar s}}} \right] + 1 + \int_0^{\bar s} {\frac{1}{{y^2 }}Put(y} )e^r dy + \int_{\bar s}^{ + \infty } {\frac{1}{{y^2 }}Call(} y)e^{rT} dy - \log (\frac{{s_T }}{{\bar s}})} \right\} \\
&= \frac{2}{T}\left\{ {rT - \frac{{s_0 e^{rT} }}{{\bar s}} + 1 - \log (\frac{{\bar s}}{{s_0 }}) + e^{rT} \int_0^{\bar s} {\frac{1}{{y^2 }}Put(y} )dy + e^{rT} \int_{\bar s}^{ + \infty } {\frac{1}{{y^2 }}Call(} y)dy} \right\} \\
\end{align*}
In this program,\[
\left\{ {\begin{array}{*{20}c}
{\ln \frac{y}{{\bar s}} \le 0,{\rm{ put}}} \\
{\ln \frac{y}{{\bar s}} > 0,{\rm{ call}}} \\
\end{array}} \right.
\]
If we define $\mbox{ }x \buildrel \Delta \over = \ln (\frac{y}{\bar
{s}}),\mbox{ them }y = e^x\cdot \bar {s}$
Then we look at
\begin{align*}
\int_0^{\bar s} {\frac{1}{{y^2 }}Put(y} )dy &= \int_{ - \infty }^0 {\frac{1}{{y^2 }}Put(y} )e^x \cdot\bar sdx \\
&= \int_{ - \infty }^0 {\frac{1}{{y^2 }}Put(y} )ydx \\
&= \int_{ - \infty }^0 {\frac{1}{y}Put(y} )dx \\
\end{align*}
Similarly ,
\[
\int_{\bar s}^{ + \infty } {\frac{1}{{y^2 }}Call(} y)dy = \int_\sigma ^{ + \infty } {\frac{1}{y}Call(} y)dx
\]
Therefore if we map the notation:
In the code : In our derivation :
\begin{center}
$\begin{array}{cc}
In the code: & In our derivation: \\
X & X \\
strike & e^x\cdot \bar{s} \\
c Strike Break Point & \bar{S}
\end{array}$
\end{center}
Then we see that the corret form should have $\frac{1}{(strike)}$ instead of
$\frac{1}{(strike)^2}$.
\section{P10}
a)\begin{align*}
E[\widehat{\beta _1 }] &= E[\frac{\sum\limits_{i = 1}^n {y_i }
}{\sum\limits_{i = 1}^n {x_i } }] \\
&= E[E[\frac{\sum\limits_{i = 1}^n {y_i } }{\sum\limits_{i = 1}^n {x_i }
}\vert x_1 , \cdots ,x_n ]] \\
&= E[E[\frac{\sum\limits_{r = 1}^n \alpha + \beta x_i + \varepsilon _i
}{\sum\limits_{i = 1}^n {x_i } }\vert x_1 , \cdots ,x_n ]] \\
&= E[n\alpha + \beta ] \\
&= n\alpha + \beta \\
\end{align*}
not unbiased !
\[
\begin{array}{l}
E[\widehat{\beta _2 }] = E[E[\frac{{\sum\limits_{i = 1}^n {y_i (x_i - \frac{1}{n}\sum\limits_{j = 1}^n {x_j )} } }}{{\sum\limits_{i = 1}^n {(x_i - \frac{1}{n}\sum\limits_{j = 1}^n {x_j )^2 } } }}|x_1 , \cdots ,x_n ]] \\
= E[E[\frac{{\sum\limits_{i = 1}^n {(\alpha + \beta x_i + \varepsilon _i )(x_i - \frac{1}{n}\sum\limits_{j = 1}^n {x_j )} } }}{{\sum\limits_{i = 1}^n {(x_i - \frac{1}{n}\sum\limits_{j = 1}^n {x_j )^2 } } }}|x_1 , \cdots ,x_n ]] \\
= E[E[\frac{{\beta \sum\limits_{i = 1}^n {(x_i - \bar x)^2 } }}{{\sum\limits_{i = 1}^n {(x_i - \bar x)^2 } }}|x_1 , \cdots ,x_n ]] \\
= \beta \\
\end{array}
\]
Unbiased !
\[
\begin{array}{l}
E[\widehat{\beta _3 }] = E[E[\frac{\sum\limits_{i = 1}^n {y_i x_i }
}{\sum\limits_{i = 1}^n {x_i^2 } }\vert x_1 , \cdots ,x_n ]] \\
= E[E[\frac{\sum\limits_{i = 1}^n {(\alpha + \beta x_i + \varepsilon _i
)x_i } }{\sum\limits_{i = 1}^n {x_i^2 } }\vert x_1 , \cdots ,x_n ]] \\
= \alpha E[\sum\limits_{i = 1}^n {x_i } ] + \beta \\
\end{array}
\]
Not unbiased !
\[
E[\widehat{\beta _4 }] = E[\frac{\sum\limits_{i = 1}^n {z_i (y_i - \overline
y )} }{\sum\limits_{i = 1}^n {(z_i - \overline z )^2x_i } }]
\]
We need more information about $z{\_}i$ in order to judge whether it's
unbiased or not.
\noindent
b) $\widehat{\beta _2 }$ha the smallest variance.(we only had one choice)
\noindent
c) When $x_i $ is correlated with $\varepsilon _i $, that means the
error/residual still has information about the predictor variables.
For example, in linear regression with time-series data, we often have $x_i
$ being correlated with $\varepsilon _i $,i.e. the disturbance has some sort
of seasonality,etc.
\noindent
d) It's called heteroscedesticity, In this case we could either put
additional predictor variables, or use GLM, GLS or weighted GLS.
\noindent
e.g.
\[
\widehat{\beta } = (x^T\varphi ^{ - 1}x)^{ - 1}x^T\varphi ^{ - 1}y
\]
Where $\varphi ^{ - 1}$=weighting metrix chesen as follows
\[
\varphi ^{ - 1} = \left[ {{\begin{array}{*{20}c}
{\exp ( - x_1 r)} \hfill & \hfill & O \hfill \\
\hfill & {\exp ( - x_2 r)} \hfill & \hfill \\
O \hfill & \hfill & \ddots \hfill \\
\end{array} }} \right]
\]
Once $\beta $ is obtained , then $\alpha $ is also easy to get. The above is
based on know $\gamma $. Now we use Hildreth-lu procedure to estimate
$\gamma $. We formulate it as an uncmstra problem for $\gamma $, and search
for $\gamma $ (and corresponding $\alpha $ and $\beta $,for smallest sum of
squared residuals.
e)
\begin{align*}
y_i & = \alpha + \beta x_i + \varepsilon _i \\
\rho y_{i - 1} & = \rho \alpha + \rho \beta x_{i - 1} + \rho \varepsilon _{i
- 1} \\
y_i - \rho y_{i - 1} & = (1 - \rho )\alpha + \beta (x_i - \rho x_{i - 1} ) +
\varepsilon _i - \rho \varepsilon _{i - 1} \\
\tilde {y}_i & = \tilde {\alpha } + \tilde {\beta }(x_i - \rho x_{i - 1} ) +
v_i \\
& = \tilde {\alpha } + \tilde {\beta }\tilde {x}_i + v_i .
\end{align*}
where
\[\tilde {y}_i = y_i - \rho y_{i - 1} ,\tilde {\alpha } = (1 - \rho )\alpha
,\tilde {\beta } = \beta ,\tilde {x}_i = x_i - \rho x_{i - 1}\]
Assuming we know $\rho$, then we can transform $y_i'$s and $x_i'$s into $\tilde{y_i}'$s and $\tilde{x_i}'$s.
And therefore we could estimate $\tilde{\alpha}$ and $\tilde{\beta}$ thru OLS, and hence obtain $\alpha$ and $\beta$.
The above is based on a given known $\rho$. But we don't know $\rho$ and would like to estimate $\rho$ also.
To estimate $\rho$, we use the Hildreth-Lu procedure.
\begin{description}
\item[$\bullet$]Iterate$\rho$from -0.99 to 0.99.
\item[$\bullet$] For each $\rho$compute the residuals$\tilde{y}_i - \tilde {\alpha} - \tilde {\beta }\tilde {x}_i $ and the sum of squared residuals.
\item[$\bullet$] Choose $\rho$ and $\tilde {\alpha}$,$\tilde{\beta }$ s.t.the OLS has smallest sum of squared residuals.
\end{description}
\section{P11}
This is also called the Vasicek model. It's often used in interest rate
modeling. $\alpha $ is the rate to pull the process back to the level$\beta
$ . $\beta $ is the mean level, $k$is the volatility. The whole model is
called the mean-reverting modil.
\begin{align*}
\sigma _t^2 &= \sigma _0^2 e^{ - \alpha t} + \beta (1 - e^{ - \alpha t} ) + \int_0^t {ke^{\alpha (s - t)} } dw_s \\
\sigma _{t + T}^2 {\rm{|}}\sigma _t^2 &= \sigma _t^2 e^{ - \alpha T} + \beta (1 - e^{ - \alpha T} ) + \int_t^{t + T} {ke^{\alpha s} } dw_s \cdot e^{ - \alpha (T + t)} \\
\end{align*}
$\Rightarrow$
\begin{align*}
E\left[ {\sigma _{t + T}^2 \left| {\sigma _t^2 } \right.} \right] &= \alpha _t^2 e^{ - \alpha T} + \beta (1 - e^{ - \alpha t} ) \\
Var\left[ {\sigma _{t + T}^2 \left| {\sigma _t^2 } \right.} \right] &= e^{ - 2\alpha (T + t)} \cdot k^2 \cdot\int_t^{t + T} {e^{2\alpha s} } ds \\
&= e^{ - 2\alpha (T + t)} \cdot k^2 \cdot\frac{1}{{2\alpha }}(e^{2\alpha (T + t)} - e^{ - 2\alpha t} \\
&= \frac{{k^2 }}{{2\alpha }}(1 - e^{ - 2\alpha T} ) \\
\end{align*}
b)$\Delta (\sigma _t^2 ) = \alpha (\beta - \sigma _t^2 )\Delta t + k\sqrt {\Delta t} $
\[
\left\{ \begin{array}{l}
\begin{array}{*{20}c}
{\sigma _2^2 - \sigma _1^2 = \alpha \beta - \alpha \sigma _1^2 + kw_1 } \\
{\sigma _3^2 - \sigma _2^2 = \alpha \beta - \alpha \sigma _2^2 + kw_2 } \\
\end{array} \\
............................................. \\
\end{array} \right.
\]
\[
\Rightarrow
\]
\[
{\left\{ {\begin{array}{*{20}c}
{\sigma _2^2 = \alpha \beta - (1 - \alpha )\sigma _1^2 + kw_1 } \\
{\sigma _3^2 = \alpha \beta + (1 - \alpha )\sigma _2^2 + kw_2 } \\
\end{array}} \right.}
\]
Then we could estimate 1-$\alpha $ and them$\beta $ using OLS.
And then obtain an estimate of $k$using sample residual
c)According to my previons calculation,
\[
E\left[ {\sigma _{t + T} ^2\left| {\sigma _t ^2} \right.} \right] = \alpha
_t ^2e^{ - \alpha T} + \beta (1 - e^{ - \alpha T})
\]
Therefore \[
\hat \sigma _{t + d}^2 = e^{ - \alpha d} \cdot\sigma _t^2 - \beta e^{ - \alpha d} + \beta
\]
Where $d$ is the time horigon and it only depends on $\alpha $,$\beta $,$d$
and the most recent daily variance realigation.
Alternatively, we could also do:
\begin{align*}
\hat \sigma _{t + d}^2 &= \frac{1}{d}\sum\limits_{i = 1}^d {(\alpha _t^2 } e^{ - \alpha i} + \beta (1 - e^{ - \alpha i} )) \\
&= \frac{1}{d}(\alpha _t^2 - \beta )\sum\limits_{i = 1}^d {e^{ - \alpha i} } ) + \beta \\
&= \frac{1}{d}(\alpha _t^2 - \beta )\frac{{e^{ - \alpha } - e^{ - \alpha (d + 1)} }}{{1 - e^{ - \alpha } }} + \beta \\
\end{align*}
\begin{align*}
\hat \sigma _{t + d}^2 &= \frac{1}{d}\sum\limits_{i = 1}^d {(\alpha _t^2 } e^{ - \alpha i} + \beta (1 - e^{ - \alpha i} )) \\
&= \frac{1}{d}(\alpha _t^2 - \beta )\sum\limits_{i = 1}^d {e^{ - \alpha i} } ) + \beta \\
&= \frac{1}{d}(\alpha _t^2 - \beta )\frac{{e^{ - \alpha } - e^{ - \alpha (d + 1)} }}{{1 - e^{ - \alpha } }} + \beta \\
\end{align*}
\noindent
d)$\sigma _t ^2$ should be always positive, but here the model being a
Gauxssian model, it allows $\sigma _t ^2$ to be negative.
\end{document}
