<p>Let $P = \alpha A + (1-\alpha) B$ where $A$ and $B$ are returns (random) from the two assets, and $P$ is their portfolio. 
Variance of portfolio $P$ can therefore be written as</p>

<p>\begin{eqnarray*} \sigma^2_P &amp; = &amp; \alpha^2 \sigma^2_A + (1-\alpha)^2 \sigma^2_B + 2\alpha (1-\alpha)\text{Cov}(A, B) \\ &amp; \leq &amp;\alpha^2 \sigma^2_A + (1-\alpha)^2 \sigma^2_B + 2\alpha (1-\alpha)\sigma_A\sigma_B \\ &amp; = &amp;(\alpha\sigma_A + (1-\alpha)\sigma_B)^2 \\ &amp; \leq &amp; (\alpha \max(\sigma_A, \sigma_B) + (1-\alpha)\max(\sigma_A, \sigma_B))^2 \\ &amp; = &amp; \max(\sigma_A^2, \sigma_B^2)\end{eqnarray*}</p>

CC BY-SA 3.0
2017-05-26T05:23:28.643Z
2017-05-26T05:23:28.643Z
Amit
<p>Let $w$ denote the weight on $A$ so that $1-w$ is the weight on $B$. 
Recall from the properties of variance that</p>

<p>$\sigma_p^2 = w^2\sigma_A^2  + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2$</p>

<p>Without loss of generality, assume $\sigma_A \geq \sigma_B$.  We wish to show that </p>

<p>$w^2\sigma_A^2  + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2\leq \sigma_A^2$</p>

<p>Note that </p>

<p>$\sigma_A^2 = \sigma_A^2 (w + (1-w)) ^2 = \sigma_A^2 w^2 + 2w(1-w)\sigma_A^2 + \sigma_A^2(1-w)^2$</p>

<p>Since $\sigma_A \geq \sigma_B$ and $w$, $(1-w)$, and $\sigma_A$ are positive, this means that</p>

<p>$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2$</p>

<p>And since the correlation has the property that $-1 \leq \rho_{AB} \leq 1$ and $w$, $(1-w)$, $\sigma_B$ and $\sigma_A$ are all positive, it must be the case that</p>

<p>$\sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$</p>

<p>Therefore</p>

<p>$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$ $\square$</p>

<p>In words, looking at the formula for variance of a convex combination of random variables, the variance is maximized if the correlation between the assets is 1.  In this case, the possible portfolio values as a function of $w$ are a straight line segment between $A$ and $B$, which clearly can't have a variance higher than either.  Now, if the correlation is less than 1, then any combination of the two will be lower than the straight line case.</p>

<p>Intuitively, the returns to assets $A$ and $B$ will partially cancel each other out any time they are not a fixed multiple of each other.  This canceling out behavior reduces the variance of the resulting portfolio.  The worst-case scenario is that the two assets are equal to each other, so the portfolio can never have a higher variance than the component asst with the highest variance.</p>

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2017-05-26T14:12:06.023Z
Doesn't this hold iff $${\sigma _A>\sigma_B}$$ ?
Jaffar 
CC BY-SA 3.0
2017-05-26T15:20:18.423Z
Without loss of generality.  If $\sigma_A=\sigma_B$, the above holds as is. If $\sigma_B \geq \sigma_A$ then switch all A and B.
farnsy
CC BY-SA 3.0
2017-05-26T05:51:06.067Z
2017-05-26T15:18:24.187Z
2017-05-26T15:18:24.187Z
farnsy
<p>When reading on Markowitz's portfolio theory, I stumbled across the fact that in a market with two risky assets, if no short selling is not allowed, the variance of a portfolio consisting of the two assets cannot exceed the variances of the risky assets individually. That is:</p>

<p>$${\sigma _p}^2 \le \max \{ {\sigma _A}^2,{\sigma _B}^2\} $$
Where A and B are two different assets.</p>

<p>Could you kindly prove this statement, and possible provide some intuition for why this is the case?</p>

CC BY-SA 3.0
2017-05-28T04:47:18.073Z
This question has received some votes to be closed because it is perhaps off topic. However, finance is definitely on-topic here. This, however, comes with the caveat that some questions are better suited to quant.SE. See these two meta posts: https://economics.meta.stackexchange.com/questions/32/is-finance-under-our-domain/34#34
jmbejara
CC BY-SA 3.0
2017-05-28T04:47:21.543Z
https://economics.meta.stackexchange.com/questions/1341/might-basic-finance-questions-be-on-topic
jmbejara
<mathematical-economics><finance><financial-economics>
Prove that variance of a portfolio cannot exceed variance of individual assets
Jaffar 
# Prove that variance of a portfolio cannot exceed variance of individual assets

(Asked by: username_0 on 2017-05-26)

When reading on Markowitz's portfolio theory, I stumbled across the fact that in a market with two risky assets, if no short selling is not allowed, the variance of a portfolio consisting of the two assets cannot exceed the variances of the risky assets individually. That is:

$${\sigma \_p}^2 \le \max \{ {\sigma \_A}^2,{\sigma \_B}^2\} $$
Where A and B are two different assets.

Could you kindly prove this statement, and possible provide some intuition for why this is the case?

**Tags:** mathematical-economics, finance, financial-economics

> **Comments:**
> **username_1**: This question has received some votes to be closed because it is perhaps off topic. However, finance is definitely on-topic here. This, however, comes with the caveat that some questions are better suited to quant.SE. See these two meta posts: https://economics.meta.stackexchange.com/questions/32/is-finance-under-our-domain/34#34  
> **username_1**: https://economics.meta.stackexchange.com/questions/1341/might-basic-finance-questions-be-on-topic  

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## 2 answers

### Answer by username_3 on 2017-05-26. Score: 5 | Accepted answer

Let $w$ denote the weight on $A$ so that $1-w$ is the weight on $B$.
Recall from the properties of variance that

$\sigma\_p^2 = w^2\sigma\_A^2 + 2w(1-w)\sigma\_A\sigma\_B \rho\_{AB}+ (1-w)^2\sigma\_B^2$

Without loss of generality, assume $\sigma\_A \geq \sigma\_B$. We wish to show that

$w^2\sigma\_A^2 + 2w(1-w)\sigma\_A\sigma\_B \rho\_{AB}+ (1-w)^2\sigma\_B^2\leq \sigma\_A^2$

Note that

$\sigma\_A^2 = \sigma\_A^2 (w + (1-w)) ^2 = \sigma\_A^2 w^2 + 2w(1-w)\sigma\_A^2 + \sigma\_A^2(1-w)^2$

Since $\sigma\_A \geq \sigma\_B$ and $w$, $(1-w)$, and $\sigma\_A$ are positive, this means that

$\sigma\_A^2 \geq \sigma\_A^2 w^2 + 2w(1-w)\sigma\_A\sigma\_B + \sigma\_B^2(1-w)^2$

And since the correlation has the property that $-1 \leq \rho\_{AB} \leq 1$ and $w$, $(1-w)$, $\sigma\_B$ and $\sigma\_A$ are all positive, it must be the case that

$\sigma\_A^2 w^2 + 2w(1-w)\sigma\_A\sigma\_B + \sigma\_B^2(1-w)^2 \geq \sigma\_A^2 w^2 + 2w(1-w)\sigma\_A\sigma\_B\rho\_{AB} + \sigma\_B^2(1-w)^2$

Therefore

$\sigma\_A^2 \geq \sigma\_A^2 w^2 + 2w(1-w)\sigma\_A\sigma\_B\rho\_{AB} + \sigma\_B^2(1-w)^2$ $\square$

In words, looking at the formula for variance of a convex combination of random variables, the variance is maximized if the correlation between the assets is 1. In this case, the possible portfolio values as a function of $w$ are a straight line segment between $A$ and $B$, which clearly can't have a variance higher than either. Now, if the correlation is less than 1, then any combination of the two will be lower than the straight line case.

Intuitively, the returns to assets $A$ and $B$ will partially cancel each other out any time they are not a fixed multiple of each other. This canceling out behavior reduces the variance of the resulting portfolio. The worst-case scenario is that the two assets are equal to each other, so the portfolio can never have a higher variance than the component asst with the highest variance.

> **Comments:**
> **username_0**: Doesn't this hold iff $${\sigma _A>\sigma_B}$$ ?  
> **username_3**: Without loss of generality.  If $\sigma_A=\sigma_B$, the above holds as is. If $\sigma_B \geq \sigma_A$ then switch all A and B.  

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### Answer by username_2 on 2017-05-26. Score: 7

Let $P = \alpha A + (1-\alpha) B$ where $A$ and $B$ are returns (random) from the two assets, and $P$ is their portfolio.
Variance of portfolio $P$ can therefore be written as

\begin{eqnarray\*} \sigma^2\_P & = & \alpha^2 \sigma^2\_A + (1-\alpha)^2 \sigma^2\_B + 2\alpha (1-\alpha)\text{Cov}(A, B) \\ & \leq &\alpha^2 \sigma^2\_A + (1-\alpha)^2 \sigma^2\_B + 2\alpha (1-\alpha)\sigma\_A\sigma\_B \\ & = &(\alpha\sigma\_A + (1-\alpha)\sigma\_B)^2 \\ & \leq & (\alpha \max(\sigma\_A, \sigma\_B) + (1-\alpha)\max(\sigma\_A, \sigma\_B))^2 \\ & = & \max(\sigma\_A^2, \sigma\_B^2)\end{eqnarray\*}


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