# CVaR formulation

I am a research intern and I am working on a topic about a profit maximization of a risk-averse newsvendor by using Conditional Value-at-Risk.The problem is that I found different expressions of CVaR. In a risk-averse newsboy problem papers, I have found the following formula :

But, in risk management papers (finance etc), I have found the following one with its proof :

The first formula is a maximization problem and the second one, it is a minimizarion.

The problem is that I coudn'd find the link between the two formulas.

π(μ,D) : π is a profit function which depends on some factors that we can control μ (decision variables vector) and D represents randomness and this case it is random demand. Y : is a random variable that represents loss function. α is variable. It does not have a special signification. But we can prove that Value-at-Risk is a solution of the second optimization problem. I thing there is something missing but I dont know what because first we talk about profit and then we talk aboout loss. Maybe there is something missing related to this.

Thanks

• I would suggest you edit the question by adding more context and definitions for all notations. May 29, 2019 at 13:17
• Keep in mind that maximization of $x$ is equivalent to minimization of $-x$. So it is important to understand what $a$ and $\nu$ represent. For example are we trying to maximize a gain or minimize a loss. And do we lose when $\{\cdots\}$ is too high or do we lose when it is too low. May 29, 2019 at 14:27
• Thanks for the edit, but still not clear. Please provide definitions to all notations. For example, what is $\mu$, what is $D$, and what is $\pi(\mu, D)$? Similarly, what are $Y$ and $\alpha$? May 29, 2019 at 14:53
• π(μ,D) : π is a profit function which depends on some factors that we can control μ (decision variables vector) and D represents randomness and this case it is random demand. Y : is a random variable that represents loss function. α is variable. It does not have a special signification. But we can prove that Value-at-Risk is a solution of the second optimization problem. I thing there is something missing but I dont know what because first we talk about profit and then we talk aboout loss. Maybe there is something missing related to this. I hope it's clearer this time. Thanks May 29, 2019 at 19:08
• $Y$ and $\pi(\mu, D)$ must have some relationships to make the comparison meaningful- you can not just comparing two random variables. Please edit directly in your question. May 29, 2019 at 19:17

If $$Y=-\pi(\mu,D)$$ then the first formula is $$\mathrm{CVaR}_\eta(-Y)=\max_{\nu\in R}\left\{\nu+\frac1\eta E((-Y-\nu)^-)\right\}$$ where $$X^-=\min (X,0)$$ and $$X^+=\max(X,0)$$. Note that $$(-X)^-=-(X^+)$$.

If we let $$1-\alpha=\eta$$ and $$\nu=-a$$ this becomes (assuming $$\max=\sup$$, i.e. the sup is attained, and using $$\sup(\mathcal A)=-\inf(-\mathcal A)$$): $$\begin{eqnarray*}\max_{\nu\in R}\left\{\nu+\frac{1}{1-\alpha} E(-((Y+\nu)^+))\right\}&=&\sup_{\nu\in R}\left\{\nu+\frac{-1}{1-\alpha} E((Y+\nu)^+)\right\}\\ =\sup_{a\in R}\left\{-a+\frac{-1}{1-\alpha} E((Y-a)^+)\right\} &=&-\inf_{a\in R}\left\{-\left(-a+\frac{-1}{1-\alpha} E((Y-a)^+)\right)\right\}\\ &=&-\inf_{a\in R}\left\{a+\frac{1}{1-\alpha} E((Y-a)^+)\right\}\end{eqnarray*}$$

Now let's imagine $$\pi(\mu,D)$$ is profit and $$Y=-\pi(\mu,D)$$ is a corresponding loss.

So the CVaR of $$Y$$, the loss, according to the second formula, is the negative of the CVaR of the profit, $$-Y$$, according to the first formula.

So I guess when dealing with a loss we take the CVaR to be negative, see e.g. an answer by Kozarevic.

• Thanks. What you did sound correct but personaly, I stil missing something. In the case of a profit function, let call it Y, CVaR is the expected profit if Y exceeds or doesnt exceed VaR(Y). If Y is a loss function, CVaR is the expected loss if Y exceeds or doesnt exceed VaR. I hope my question is clear. Thanks I am really confused between loss and profit functions and how VaR and CVaR expressions vary with respect to if Y is a loss or a profit function. May 29, 2019 at 20:52
• @MohamedChaaban OK I added something, maybe it helps May 30, 2019 at 0:31
• @kjos-hanssen thanks a lot man. May 30, 2019 at 2:36