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 ?

  • $\begingroup$ I have VAR for commodity 1 and VAR for commodity 2. $\endgroup$ Commented Nov 23, 2016 at 0:48
  • $\begingroup$ What do you mean? Do you want to compute the VaR of holding 1 unit of both commodities? $\endgroup$
    – SRKX
    Commented Nov 23, 2016 at 1:15
  • $\begingroup$ Yes var of holding both $\endgroup$ Commented Nov 23, 2016 at 1:16

4 Answers 4


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:

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


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

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:

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

  • $\begingroup$ Normal distribution is very rare situation in practice. Is it possible to extend your answer to general case? $\endgroup$
    – Nick
    Commented Nov 23, 2016 at 7:23
  • $\begingroup$ @Nick general case has no simple closed-form solution; it would depend on the joint distribution of the assets. $\endgroup$
    – SRKX
    Commented Nov 23, 2016 at 7:36
  • $\begingroup$ what is "w" in the above? $\endgroup$ Commented Nov 23, 2016 at 12:58
  • $\begingroup$ I'm guessing w is weight of % of each i own $\endgroup$ Commented Nov 23, 2016 at 13:04
  • $\begingroup$ Yes that's correct. And you VaR is expressed as a percentage of your initial wealth. $\endgroup$
    – SRKX
    Commented Nov 23, 2016 at 14:10

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

  • $\begingroup$ That's it? I was reading about it violating sub additivity $\endgroup$ Commented Nov 23, 2016 at 1:07
  • $\begingroup$ VAR is the abbreviation of variation, VaR is the abbreviation value at risk. In the title you wrote value at risk, in the body you used VAR. What did you mean? $\endgroup$
    – Nick
    Commented Nov 23, 2016 at 3:37
  • $\begingroup$ Value at risk yes. Sorry about abbreviation. Value at risk in title. $\endgroup$ Commented Nov 23, 2016 at 4:35
  • $\begingroup$ @Nick I think VAR is more commonly used as Vector AutoRegression, though. $\endgroup$
    – Jan Sila
    Commented Nov 23, 2016 at 8:31

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.


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?

  • $\begingroup$ This is only showing it would not work in all cases, but you don't show why it couldn't work under certain assumption (here $\rho=0$). $\endgroup$
    – SRKX
    Commented Nov 23, 2016 at 5:49

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