Absolutely! Let's break down the problem step by step: --- **2. Log-Returns with same mean and variance, but from different distributions** You’re given a stock price model: \[ S_T = S_0 e^{(\mu - \sigma^2/2)T + \sigma \sqrt{T} W} \] where: - \(W\) is a random variable with mean 0 and variance 1. - You have two cases: \(W = Y\) (some distribution from the previous question) and \(W = Z\) (standard normal \(\mathcal{N}(0,1)\)). - The parameters: \(S_0 = 50\), \(\mu = 0.2\), \(\sigma = 0.6\), \(T = 0.5\). You consider a position that is long 1 unit of the stock, and you are interested in the **loss**: \[ L_T = -(S_T - S_0) = S_0 - S_T \] Let's answer each part: --- ## (a) Compute the cdf and pdf of the loss \(L_T\) for both cases **Hint:** Express \(L_T\) in terms of \(W\), then use the distribution of \(W\). ### Step 1: Express \(L_T\) in terms of \(W\) \[ S_T = S_0 e^{(\mu - \sigma^2/2)T + \sigma \sqrt{T} W} \] \[ L_T = S_0 - S_T = S_0 - S_0 e^{(\mu - \sigma^2/2)T + \sigma \sqrt{T} W} \] \[ L_T = S_0 \left[1 - e^{(\mu - \sigma^2/2)T + \sigma \sqrt{T} W}\right] \] ### Step 2: Find the cdf and pdf Let \(X = (\mu - \sigma^2/2)T + \sigma \sqrt{T} W\). Then, \[ L_T = S_0 (1 - e^X) \] Given \(W\), \(X\) is a monotonic function of \(W\). #### CDF: \[ F_{L_T}(l) = P(L_T \leq l) = P(S_0 (1 - e^X) \leq l) \] \[ \Rightarrow 1 - e^X \leq \frac{l}{S_0} \] \[ \Rightarrow e^X \geq 1 - \frac{l}{S_0} \] \[ \Rightarrow X \geq \ln\left(1 - \frac{l}{S_0}\right) \] Since \(X\) is a linear function of \(W\), for \(W = Y\) or \(Z\): \[ X = (\mu - \sigma^2/2)T + \sigma\sqrt{T}W \] \[ W \geq \frac{\ln(1 - l/S_0) - (\mu - \sigma^2/2)T}{\sigma\sqrt{T}} \] Let’s call this threshold \(w^*\): \[ w^* = \frac{\ln(1 - l/S_0) - (\mu - \sigma^2/2)T}{\sigma\sqrt{T}} \] So, \[ F_{L_T}(l) = P(W \geq w^*) \] or \[ F_{L_T}(l) = 1 - F_W(w^*) \] where \(F_W\) is the cdf of \(W\). #### PDF: \[ f_{L_T}(l) = \frac{d}{dl} F_{L_T}(l) = -f_W(w^*) \cdot \frac{d w^*}{dl} \] Compute the derivative: \[ \frac{d w^*}{dl} = \frac{-1}{\sigma\sqrt{T}(1 - l/S_0) S_0} \] So, \[ f_{L_T}(l) = f_W(w^*) \cdot \frac{1}{\sigma\sqrt{T}(1 - l/S_0) S_0} \] **In summary:** - For both \(W = Y\) and \(W = Z\), use their respective cdf/pdf in the above formulas. --- ## (b) Plot the pdfs and compare - On the same axis, plot the pdfs \(f_{L_T}(l)\) for \(W = Y\) and \(W = Z\). - **Brief comparison:** The shape of the loss distributions will depend on the tail behavior of \(Y\) and \(Z\). If \(Y\) has heavier tails, the loss distribution will also have heavier tails, indicating higher probability of extreme losses. --- ## (c) Derive the quantile function of \(L_T\) Quantile function: Given \(q\), find \(l_q\) such that \(P(L_T \leq l_q) = q\). From above: \[ F_{L_T}(l_q) = 1 - F_W(w^*) = q \] \[ F_W(w^*) = 1 - q \] \[ w^* = F_W^{-1}(1 - q) \] Recall: \[ l_q = S_0 (1 - e^{(\mu-\sigma^2/2)T + \sigma\sqrt{T}w_q}) \] where \(w_q = F_W^{-1}(1 - q)\). --- ## (d) Compute the Value-at-Risk (VaR) at 95%, 99%, and 99.9% - **VaR at level \(q\):** This is just the quantile \(l_q\) for \(q = 0.95, 0.99, 0.999\). - **Compare:** For heavier-tailed \(Y\), VaR will be larger at high confidence levels (e.g., 99%, 99.9%), reflecting larger potential for extreme losses. --- ### **Summary Table** | Step | Formula | |------|---------| | Loss | \(L_T = S_0 (1 - e^{(\mu-\sigma^2/2)T + \sigma\sqrt{T}W})\) | | CDF | \(F_{L_T}(l) = 1 - F_W(w^*)\), \(w^* = \frac{\ln(1 - l/S_0) - (\mu-\sigma^2/2)T}{\sigma\sqrt{T}}\) | | PDF | \(f_{L_T}(l) = f_W(w^*) / [\sigma\sqrt{T} (1 - l/S_0) S_0]\) | | Quantile | \(l_q = S_0 [1 - e^{(\mu-\sigma^2/2)T + \sigma\sqrt{T} w_q}]\), \(w_q = F_W^{-1}(1 - q)\) | | VaR | \(VaR_q = l_q\) | --- If you need help with plotting or coding, or the specifics about the distribution \(Y\), let me know!