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2

This will be the inverse process $$\frac{1}{S_t}$$ Applying Itô's formula the dynamics are then given by $$d\frac{1}{S_t}=\frac{-1}{S_t^2}dS_t+\frac{1}{S_t^3}dS_tdS_t$$ some simple algebra then leads to $$d\frac{1}{S_t}=\frac{1}{S_t}(\sigma^2 -r)dt+\frac{1}{S_t}\sigma dW_t$$

4

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.

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