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The way I calculate it is summing up the weighted beta of long and shorts but I saw a table where this wasn't the case so I am wondering if this is not the correct way.

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    $\begingroup$ Can you show the table and tell us where it comes from? $\endgroup$
    – SRKX
    Commented Jul 1, 2016 at 2:05
  • $\begingroup$ The source of the table is what I don't have. It's just numbers stating Beta of Long = 1.2 , beta of short = -0.85 and net = 0.6 which isn't the sum of it. $\endgroup$
    – Vandana
    Commented Jul 1, 2016 at 13:16
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    $\begingroup$ Well then, doesn't that just mean that the weight of the long portfolio and the weight of the short portfolio are not [0.50 0.50] , but rather you have more dollars in your long portfolio than in your short? In other words they calculated betas separately for the long portfolio and the short portfolio as if they were independent portfolios. $\endgroup$
    – Alex C
    Commented Jul 2, 2016 at 13:03
  • $\begingroup$ You on the other hand computed the beta for a single portfolio. The final result (the overall beta) should be the same. $\endgroup$
    – Alex C
    Commented Jul 2, 2016 at 13:26

1 Answer 1

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You can actually show by construction that the beta of the portfolio is the weighted sum of all the underlyings betas.

Assume the return of the benchmark and some asset $a$ at time $t$ are respectively denoted $r_{b,t}$ and $r_{a,t}$, then the beta of a given asset is defined by:

$$r_{a,t} = \alpha_a + \beta_a r_{b,t} + \epsilon_{a,t}$$

Let's assume you have a portfolio of $n$ assets $(a_1, ..., a_i, ..., a_n)$ each with a weight $w_i$, then the return of the portfolio is at time $t$ is defined as:

$$r_{p,t} = \sum_{i=1}^n w_i r_{a_i,t}$$

Now, by expressing each asset's return in terms of their own beta, you get:

$$ \begin{align} r_{p,t} &= \sum_{i=1}^n w_i r_{a_i,t}\\ &= \sum_{i=1}^n w_i \left( \alpha_{a_i} + \beta_{a_i} r_{b,t} + \epsilon_{a_i,t} \right)\\ &= \underbrace{\sum_{i=1}^n w_i \alpha_{a_i}}_{\alpha_p} + \underbrace{\left(\sum_{i=1}^n w_i \beta_{a_i} \right)}_{\beta_p} r_{b,t} +\sum_{i=1}^n w_i \epsilon_{a_i,t} \end{align} $$

There might be something in the table that you missed (likely the weights as Alex C pointed out) or maybe it was wrong.

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  • $\begingroup$ It does look like a weights issue. Thanks for looking into this. $\endgroup$
    – Vandana
    Commented Jul 5, 2016 at 13:16

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