Why Standalone Value at Risk Estimates Usually Cannot Be Added

# Is Value At Risk additive?


# Is Value At Risk additive?












I have computed the value at risk of 2 different commodities.

Assuming they have not correlated, can I just sum the two standalone VaR to get my overall portfolio's VaR ?

## Answer by SRKX (score 7, accepted)

https://quant.stackexchange.com/a/31145

The answer to your question is no. Value at Risk is not additive in the sense that $\text{VaR}(X+Y) \neq \text{VaR}(X) + \text{VaR}(Y)$. But I guess your question is more to aimed at finding a formula for your investments than to look at the property itself.

I think the only way to get a nice formula for this is to assume that both assets are:

- Normally distributed

- Have a mean equal to 0

- Are independent

Closed-Form value at risk for Normal variable

Mathematically the Value at Risk at a given level $\alpha$ is defined as:

$$\text{VaR}_\alpha(X) = \{ y ~ | ~ \mathbb{P}( X\leq y) = \alpha \}$$

If you can assume you variable $X$ is normally distributed such that $X \sim \mathcal{N}(\mu, \sigma^2)$, then you can re-express $X$ in terms of another variable $Z \sim \mathcal{N}(0,1)$: $X = \mu + \sigma Z$.

Using this, we know can rewrite the VaR definition as:

$$\begin{align} \text{VaR}_\alpha(X) &= \{ y ~ | ~ \mathbb{P}( \mu + Z\sigma\leq y) = \alpha\}\\ &= \left\{ y ~ | ~ \mathbb{P}\left( Z \leq \frac{ y- \mu}{\sigma} \right) = \alpha \right\}\\ &= \left\{ y ~ | ~ \Phi\left( \frac{ y- \mu}{\sigma} \right) = \alpha \right\}\\ \end{align}$$

where $\Phi(x)$ is the cumulative normal standard distribution function.

We can then find a closed-form formula to the value at risk of a normally distributed variable $X$:

$$\text{VaR}_\alpha(X) = \Phi^{-1}(\alpha) \cdot\sigma + \mu$$

Distribution of portfolio of two Normal variables

Now, let's assume you portfolio $Y$ holds two assets $X_1$ and $X_2$ (the two commodities in your example), which are uncorrelated ($\rho = 0$).

If you assume that both are normally distributed $X_1 \sim \mathcal{N}(\mu_1,\sigma_1)$ and $X_2 \sim \mathcal{N}(\mu_2,\sigma_2)$, then the we know that the portfolio can be expressed as:

$$\begin{align} Y &= wX_1 + (1-w)X_2\\ &= w(\mu_1 + \sigma_1 Z_1) + (1-w)(\mu_2 +\sigma_2 Z_2)\\ &= w\mu_1 + (1-w) \mu_2 + w\sigma_1 + w \sigma_1 Z_1 + (1-w) \sigma_2 Z_2 \end{align}$$

Hence, we know that:

$$\mathbb{E}(Y) = w\mu_1 + (1-w) \mu_2$$

and

$$\text{Variance}(Y) = \sigma_Y^2 = w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2$$

because you assets are independent.

As we know, the sum of 2 normally distributed variables is also normally distributed, hence: $$Y \sim \mathcal{N}(w\mu_1 + (1-w) \mu_2, w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2)$$

Value-at-risk of the portfolio

Using the formula for value-at-risk for normal variable we found above, we can write:

$$\begin{align} \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sigma_Y + \mu_y\\ \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sqrt{w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2} + w\mu_1 + (1-w) \mu_2\\ \end{align}$$

If you assume that $\mu_1 = \mu_2 = 0$, then you get:

$$\begin{align} \text{VaR}_\alpha(Y) &= \Phi^-1(\alpha) \sqrt{w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2}\\ \text{VaR}_\alpha(Y)^2 &= \Phi^-1(\alpha)^2 (w^2 \sigma_1^2 + (1-w)^2 \sigma_2^2)\\ \text{VaR}_\alpha(Y)^2 &= \Phi^-1(\alpha)^2 w^2 \sigma_1^2 + \Phi^-1(\alpha)^2 (1-w)^2 \sigma_2^2\\ \text{VaR}_\alpha(Y)^2 &= w^2 \text{VaR}_\alpha(X_1)^2 + (1-w)^2 \text{VaR}_\alpha(X_2)^2\\ \text{VaR}_\alpha(Y) &=\sqrt{ w^2 \text{VaR}_\alpha(X_1)^2 + (1-w)^2 \text{VaR}_\alpha(X_2)^2}\\ \end{align}$$

## Answer by dm63 (score 0)

https://quant.stackexchange.com/a/31138

You need to square them, add the squares , and take the square root. (Variances are additive, not standard deviations).

## Answer by Nick (score 0)

https://quant.stackexchange.com/a/31144

No, because the value at risk is not, in general, a coherent risk measure as it does not respect the sub-additivity property, i.e.

$\rho(X + Y) \ne \rho(X) + \rho(Y)$, $\forall X, Y \in \mathcal{X}$ for the $VaR$.

However, Conditional Value at Risk is. Check out Is Conditional Value-at-Risk (CVaR) coherent?

## Answer by Luis Fernández (score 0)

https://quant.stackexchange.com/a/31154

Well, if you are using historical VaR, you can add results on each scenario and then calculate percentile of results... There is no other way.