Does the traditional NPV formula of a cashflow double count risk?

Question

Consider a cash flow stream of a single payment (1 period away). Its net present value is typically presented as

$$ \text{NPV} = {\text{EV}(\text{Cash Flow}) \over 1 + d} \tag{1} $$

Here $d$ is supposed to be the "risk-adjusted discount rate" which, as I understand it, can be broken down as

$$ d = t + r $$

where $t$ is the pure time value of money and $r$ is the "extra factor" for how risky the asset class is.

Question: Aren't we double-counting the risk of an asset by including the expected value in the numerator (which factors in the probability that a cash flow could be low) and the risk factor $r$ in the denominator? That is, shouldn't the NPV of an asset be either

$$ \text{NPV} = {\text{EV}(\text{Cash Flow}) \over 1+t} \tag{2} $$

or

$$ \text{NPV} = {\text{Cash Flow} \over 1+t + r} \tag{3}? $$

In (2) we factor in the risk of the asset by using an expected cash flow in the numerator, so that if the cash flow is really risky it will be weighted down. In (3) we factor in the risk of the asset by discounting by $1+t + r$ instead of just $t$. Crucially, we do one or the other; doing both -- as in (1) -- seems to double count the risk of an asset. How is this wrong?

Answer 1

That formula is algebraically equivalent to saying different, stochastic assets can have different expected returns.

$$ \mathbb{E} \left[ R_i \right] = r_f + \gamma_i $$

Some simple algebra

Let $X_i$ be a random variable denoting a risky cash flow, $p_i$ be today's price of that risky cash flow, $r_f$ be the risk free rate, and $\gamma_i$ be some risk premium specific to asset $i$.

The formula you're objecting to is:

$$ p_i = \frac{\mathbb{E} [X_i]}{r_f + \gamma_i}$$

Asset $i$'s return is given by $ R_i = \frac{X_i}{p_i}$ By simple algebra you get $\mathbb{E} \left[ R_i \right] = r_f + \gamma_i$.

So all that formula is saying is that the expected return of asset $i$ is the risk free rate plus some risk premium $\gamma_i$. Without the $\gamma_i$ term (which is $d$ in your notation), every asset would have to have an expected return of the risk free rate, which is obviously wrong.

Answer 2

“You can't compensate for risk by using a high discount rate." - Warren Buffett at the 1998 Berkshire Hathaway Shareholder Meeting

The simple answer to your question is, “yes, many implementations of discounted cash flow analyses which adjust the discount rate for risk are double counting”. This practice is pervasive in academia, but has no basis in the time value of money principle.

I presume this practice comes from the incorrect interpretation of the Capital Asset Pricing Model, which itself may be interpreted as an incorrect interpretation of the Modigliani-Miller postulate on the irrelevance of capital structure.

Even those who realize this continue in this practice for heuristic reasons as it approximates the intuition that NPV must be downwardly adjusted for higher risk. It also allows for the heuristic valuation of negative expected cash flows, which is intractable in a deterministic context. Moreover, adjusting for asymmetric aversion to downside risk — as laid out in Prospect Theory — is mathematically and computationally inconvenient.

To my knowledge, there is no generally accepted way to discount an annuity under a conditional probability measure, which you represent as EV(*), even when that is taken to be a real world measure (vice a risk neutral measure). The most comprehensive works in this area are by Daniel Dufresne.

Anyhow, in order to avoid going down this rabbit hole, it might be wise to continue doing things the way your supervisor or professor expects.

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Addendum, please note that @MatthewGunn 's answer is not wrong in the realm of quantitative finance in which price is assumed to be equal to the discounted expectation. I interpreted the question as a valuation/actuarial problem in which the author intends to discover fair value independently of price.

Answer 3

No, but I can tell you why it feels like you are double counting.

Consider a cash flow $\tilde{x}=\tilde{x}(t,\mu,\sigma^2)$ to be received in the future. While many cash flows lack a first moment and so no defined mean or variance, let us assume at least the second moment is defined to make the discussion simple. Implicitly, your assumption of an expectation would require a first moment to exist. Indeed to make this easier, let us assume normality.

If $t$ is time; $\mu$ the center of location; and, $\sigma^2$ the scale parameter, then we can talk about a rate. From the formula $$\rho=\frac{\mathbb{E}(\tilde{x}(\mu(t),\sigma^2(t)))}{1+d(\mu(t),\sigma^2(t))}.$$ Your assumption of additivity is problematic, while it is often used as an approximation, if you think about it for a second you will see why. Instead, I am defining $$d=(1+r(\mu_0,\sigma^2_0))(1+\tau(t))$$ because I am using $t$ for time instead of risk. $\tau(t)$ is the function that maps the premium as a function of time to discount a certain cash flow.

We will adopt a Frequentist interpretation of probability to make this simple. Using multiplication all the constants are on the left and the expectation of the random variables are on in the center when we rearrange it as $$\rho(1+d(\mu(t),\sigma^2(t))=\mathbb{E}(\tilde{x}(\mu(t),\sigma^2(t)))=\rho\mu(t).$$

The present value cannot be stochastic as it is known by observation. Since $\mu(t)$ and $\sigma^2(t)$ are constants by definition in the Frequentist interpretation of probability nothing on the left or right contains any randomness at all. Only the center is random. That randomness is averaged out over the sample space so that only the point is left.

You could logically take this one step further and argue that $\mu(t)=\mu(t,\sigma^2(t))$. It feels like it is double counted because the mean is a function of the variance and the cash flow is a function of the mean and the variance. The rate is a function of the risk-adjusted mean, so it is a function of the mean and the variance.

You could dissolve the mean and convert it into a pure function of variance and time, and then only the scale parameter would exist in the numerator and the denominator.

$$\rho=\frac{\mathbb{E}(\tilde{x}(\sigma^2(t),t))}{1+d(\sigma^2(t),t)}$$

You are also missing the observation that $\mu(t,\sigma^2(t))$ is similar to an expenditure function and not merely a center of location.

If you step back one more unit of time, to before time zero, then $\rho$ becomes stochastic as well because $\tilde{\rho}=\rho$ if and only if $\mathbb{E}[\mathcal{U}(\tilde{x})]>\mathcal{U}(\tilde{\rho}=0)$, where $\mathcal{U}()$ is a utility function.