Optimization of stop-loss contract with respect to CVaR.

Question

I'm trying to figure out where my mistake is. I'm trying to optimize a parameter for a stop-loss type contract with respect to conditional value-at-risk. I have attempted to differentiate with respect to the parameter of interest, but I'm afraid that the final expression is nonsensical. Can anyone spot a mistake in the following calculations?

Assume $X$ is a bounded continuous stochastic variable taking positive values measuring financial losses, with density function $f(x)$. We wish to hedge against losses by insuring parts of this risk: Let \begin{align} I_a(X) = \begin{cases} X, X \leq a, \\ a, X > a.\end{cases}\end{align} For this we pay the premium $\pi_a = (1+\theta)E[X - I_a(X)]$, where $\theta > 0$ is a risk premium. The aim is to find an optimal value for $a$ to minimize the conditional value-at-risk of $I_a(X) + \pi_a$; Let $\alpha \in (0, 1) $ be given, and $q_{\beta}(X) := \inf \{c \mid P(X> c)\leq \alpha \}$ be the $\alpha$-quantile of $X$. The conditional value-at-risk is defined as \begin{align} CVaR_{\alpha}(I_a(X)) := \frac{1}{1-\alpha}\int_{\alpha}^{1}I_a(q_{\beta}(X))\ d\beta. \end{align}

The goal is to solve the following optimization problem: \begin{align} \min_{a} CVaR_{\alpha}(I_a(X) + \pi_a) = \min_{a} CVaR_{\alpha}(I_a(X)) + \pi_a, \end{align} as $CVaR$ is translation invariant. I attempt to do this by differentiating this with respect to $a$. First, I note that

\begin{align} \int_{\alpha}^{1} I_a(q_{\beta}(X))\ d\beta &= \int_{q_{\alpha}}^{sup X} I_a(x)f(x)dx. \end{align}

Moreover, if $a \leq q_{\alpha}(X)$, we get that $CVaR_{\alpha}(I_a(X)) = a$. Assuming then $a > q_{\alpha}(X)$, we get:

\begin{align} \int_{\alpha}^{1} I(q_{\beta}(X))\ d\beta &= \int_{q_{\alpha}}^{a} I_a(x)f(x)dx + \int_{a}^{\sup X} a f(x)dx. \end{align} Differentiating with respect to $a$ gives:

\begin{align} \frac{d}{da} \left[ \int_{q_{\alpha}}^{a} I_a(x)f(x)dx + \int_{a}^{\sup X} a f(x)dx \right] &= I_a(a)f(a) + \frac{d}{da}\left( aP(X\geq a)\right)\\ &= a f(a) + P(X\geq a) - a f(a) \\ &= P(X\geq a), \end{align} using the fundamental theorem of calculus (and assuming continuity of $X$ and hence $I_a(X)$.)

On the other hand, \begin{align} \frac{d}{da} E[X-I_a(X)] &= \frac{d}{da} \int_{a}^{\sup X} (x-a) f(x)dx \\ &= a f(a) - af(a) -\int_{a}^{\sup X} f(x) dx \\ &= -P(X\geq a). \end{align}

Puting this together, we get:

\begin{align} \frac{d}{da} \left[ CVaR_{\alpha}(I_a(X)) + \pi_a \right] &= \frac{1}{1-\alpha} P(X\geq a) + -(1+\theta) \cdot P(X \geq a), \\ \end{align}

But this seems suspicious, as setting this equal to $0$ yields $a = \sup X$ or $\frac{1}{1-\alpha} = 1+\theta$.

Answer 1

Let $W_0$ be the initial total wealth. Insuring on minimum final wealth $W_0-a$ ($a$ is the maximum loss) means buying a cash settled put option with strike $K=W_0-a$. Indeed the final payoff is $(K-W_t)^++W_t=(W_0-a-W_t)^++W_t$ so that if $K\geq W_t$ you receive $K-W_t$ and final wealth is $K-W_t+W_t=K=W_0-a$ and if $K<W_t$ you end up with $W_t$. The price of the option is $\pi(a)$

In terms of wealth, the $\textrm{CVaR}$ of the strategy given a $\textrm{VaR}_\alpha$ is $$\textrm{CVaR}=\mathbb{E}[(K-W_t)^++W_t|(K-W_t)^++W_t\leq\textrm{VaR}_\alpha\}]$$ So clearly $$P((K-W_t)^++W_t\leq\textrm{VaR}_\alpha)=1-\alpha$$ therefore $$\textrm{CVaR}=\frac{1}{1-\alpha}\mathbb{E}[((K-W_t)^++W_t)\mathbb{I}_{\{(K-W_t)^++W_t\leq \textrm{VaR}_\alpha\}}]$$ If $\textrm{VaR}_\alpha\leq K$ then $\textrm{CVaR}=W_0-a$ otherwise $$(1-\alpha)\textrm{CVaR}=(W_0-a)P(W_t\leq K)+\int_{W_0-a}^{\textrm{VaR}_\alpha}wf_{W_t}(w)dw$$ Derivative wrt $a$: $$-W_0f_{W_t}(W_0-a)-P(W_t\leq W_0-a)+af_{W_t}(W_0-a)+(W_0-a)f_{W_t}(W_0-a)=-P(W_t\leq W_0-a)$$ which is equivalent to your result $-P(X\geq a)$. Now we want to change $a$ to minimize the $\textrm{CVaR}+\pi(a)$, by taking the derivative and putting it equal to $0$: $$-\frac{P(W_t\leq W_0-a)}{1-\alpha}+\frac{\partial}{\partial a}\pi(a)=0$$ But we know that $$\pi(a)=m\mathbb{E}[(K-W_t)^+]$$ So $$\frac{\partial}{\partial a}\pi(a)=mP(W_t\leq W_0-a) \implies m(1-\alpha)=1$$ Which does look suspicious and it is equal to your result. So my guess is that the optimization setting is flawed and you must choose a loss function to minimize (like a quadratic loss or something).

Answer 2

To answer Snoop's comment:

Yes, I think in terms of wealth if VaRα≤W0−a then we get P(Wt≤W0−a)=1m and we choose the a to obtain such equation, and equivalently in terms of loss we get a VaR like you said. However the issue might be taking the derivative of the function to be optimized. Have you checked what happens with numerical optimization?

Using Monte Carlo sampling with $X$ having an exponential distribution with $\lambda = 30$, the following plots show the $CVaR(I_a(X) + \pi_a(X))$ with $\theta = 1.1, \alpha = 0.95.$

In the plot to the right, the value $a = VaR_{1-1/(1+\theta)}(X) = VaR_(0.0654)(X) = 2.029$ is indicated, and optimal for this problem. $VaR_{0.95}(X) = 91.79$. For $a \in (0, 91.79]$ The optimal value is then the indicated optimum, and for $a > 91.79$ the loss is maximized at $a = sup(X)$, with no local minimum on the interval $(91.79, \sup(X)]$. I wonder if this supports the conclusion from the original post.