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If the loss distribution is normal with mean $\mu$ and variance $\sigma^2$, then the Value-at-Risk and Expexted Shortfall (or CVaR) at level $\alpha \in (0, 1)$ are \begin{align*} \mbox{VaR}_\alpha & = \mu + \sigma \Phi^{-1}(\alpha) , \\ \mbox{ES}_\alpha & = \mu + \sigma \frac{\phi\{\Phi^{-1}(\alpha)\}}{1 - \alpha} , \end{align*} where $\phi$ ...


3

A key property of Brownian motion is independent increments. So if $x-1 > y$, then $$ \mathbb{E}[\Delta W_x \Delta W_y] = 0 $$ because the time intervals [x-1,x] and [y-1,y] do not overlap. If they do overlap, i.e. $x-1 \leq y < x$, then \begin{align} \mathbb{E}[\Delta W_x \Delta W_y] =&\ \mathbb{E}[(W_x - W_{x-1}) (W_y-W_{y-1})] \\ =&\ ...


2

I don't know what you did when you tried pulling out $1-\alpha$, the correct expression would be $\lim_{\alpha \to 1} \frac{\mu(1-\alpha) + \sigma {\phi^{-1}(\alpha)}}{(1-\alpha)(\mu + \sigma \phi^{-1}(\alpha))}$. Anyhow, you can try using the substitution $\Phi^{-1}(\alpha) = x$, $x \to \infty$ and $\alpha = \Phi(x)$. Then the expression becomes ...


3

The initial condition for the backward Kolmogorov PDE is that $$ u(0,x) = g(x) $$ for all $x$ in the relevant domain and not just at a particular point. So if your functions $f$ and $g$ agree only at a single point the initial conditions are in fact different.


3

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ is known and fixed and look at by Ito's formula $$ d(S_t/S_0) = \mu dt + \sigma B_t + \frac{\sigma^2}{2} dt. $$ Then with some abuse of notation: $$ ...


4

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[e^{\sigma W}]> e^{\sigma E[W]}=1$. $\sigma$ can be interpreted as the magnitude of the convexity of the exponential function. This can be seen by Taylor ...


2

$$S_t = S_0\exp((r-\frac{\sigma^2}{2})t+\sigma W_t)$$ is not yet a martingale for it is not dirftless. From a probabilistic point of vew the "drift adjustment" comes into play so that the expected value of $S_t$ will be $e^{rt}$ rathern than $e^{(r+0.5\sigma^2)t}$. For the expected value of a log-normaly distributed variable with mean $\mu$ and vol ...



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