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[asset_pricing_lph] Adjustments for new lecture (#98)
* display math own lines, move figure to _static * removing \ from math Co-authored-by: AakashGC <aakashgfude@gmail.com>
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lectures/AssetPricing_v1.jpg renamed to lectures/_static/lecture_specific/asset_pricing_lph/AssetPricing_v1.jpg

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lectures/asset_pricing_lph.md

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jupytext:
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extension: .md
@@ -46,16 +46,18 @@ We begin with a **key asset pricing equation**:
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$$
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E m R^i = 1
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E m R^i = 1
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$$ (eq:EMR1)
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for $i=1, \ldots, I$ and where
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$$\begin{aligned}
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$$
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\begin{aligned}
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m &=\text { stochastic discount factor } \\
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R^{i} &= \text {random gross return on asset } i \\
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E &\sim \text { mathematical expectation }
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\end{aligned}$$
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\end{aligned}
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$$
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The random gross returns $R^i$ and the scalar stochastic discount factor $m$ live
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live in a common probability space.
@@ -87,7 +89,9 @@ $\operatorname{cov}\left(m, R^{i}\right) = E (m - E m)(R^i - E R^i) $
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of a
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covariance implies that
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$$ E m R^i = E m E R^{i}+\operatorname{cov}\left(m, R^{i}\right) $$
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$$
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E m R^i = E m E R^{i}+\operatorname{cov}\left(m, R^{i}\right)
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$$
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Substituting this result into
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key equation {eq}`eq:EMR1` gives
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which can be rearranged to become
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$$E R^i = R^{f}-\operatorname{cov}\left(m, R^{i}\right) R^{f} . $$
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$$
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E R^i = R^{f}-\operatorname{cov}\left(m, R^{i}\right) R^{f} .
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$$
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It follows that we can express an **excess return** $E R^{i}-R^{f}$ on asset $i$ relative to the risk-free rate as
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$$ E R^{i}-R^{f} = -\operatorname{cov}\left(m, R^{i}\right) R^{f}
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$$
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E R^{i}-R^{f} = -\operatorname{cov}\left(m, R^{i}\right) R^{f}
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$$ (eq:EMR3)
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@@ -135,7 +142,7 @@ Equation {eq}`eq:EMR3` can be rearranged to display important parts of asset pri
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We can obtain the celebrated **expected-return-Beta -representation** for gross return $R^i$ simply by rearranging excess return equation {eq}`eq:EMR3` to become
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$$
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E R^{i}=R^{f}+\left(\underbrace{\frac{\operatorname{cov}\left(R^{i}, m\right)}{\operatorname{var}(m)}}_{\quad\quad\beta_{i,m} = \\ \text{regression coefficient}}\right)\left(\underbrace{-\frac{\operatorname{var}(m)}{E(m)}}_{\quad\lambda_{m} = \\ \text{price of risk}}\right)
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E R^{i}=R^{f}+\left(\underbrace{\frac{\operatorname{cov}\left(R^{i}, m\right)}{\operatorname{var}(m)}}_{\quad\quad\beta_{i,m} = \text{regression coefficient}}\right)\left(\underbrace{-\frac{\operatorname{var}(m)}{E(m)}}_{\quad\lambda_{m} = \text{price of risk}}\right)
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$$
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or
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A popular model of $m$ is
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$$ m_{t+1} = \exp(-\rho) \exp(- \gamma(c_{t+1} - c_t)) $$
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$$
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m_{t+1} = \exp(-\rho) \exp(- \gamma(c_{t+1} - c_t))
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$$
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where $ \rho > 0$, $\gamma > 0$, and the log of consumption growth is governed by
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$$ c_{t+1} - c_t = \mu + \sigma_c \epsilon_{t+1} $$
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$$
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c_{t+1} - c_t = \mu + \sigma_c \epsilon_{t+1}
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$$
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where $\epsilon_{t+1} \sim {\mathcal N}(0,1)$.
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@@ -173,15 +184,21 @@ Here
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$$ m_{t+1} = \exp(-\rho) \exp( - \gamma \mu - \gamma \sigma_c \epsilon_{t+1}) $$
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$$
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m_{t+1} = \exp(-\rho) \exp( - \gamma \mu - \gamma \sigma_c \epsilon_{t+1})
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$$
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In this case
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$$ E m_{t+1} = \exp(-\rho) \exp \left( - \gamma \mu + \frac{\sigma_c^2 \gamma^2}{2} \right) $$
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$$
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E m_{t+1} = \exp(-\rho) \exp \left( - \gamma \mu + \frac{\sigma_c^2 \gamma^2}{2} \right)
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$$
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and
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$$ \operatorname{var}(m_{t+1}) = E(m) [ \exp(\sigma_c^2 \gamma^2) - 1) ] $$
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$$
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\operatorname{var}(m_{t+1}) = E(m) [ \exp(\sigma_c^2 \gamma^2) - 1) ]
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$$
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When $\gamma >0$, it is true that
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@@ -224,15 +241,18 @@ $$ (eq:EMR5)
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where $\rho_{m, R^i}$ is the correlation coefficient defined as
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$$ \rho_{m, R^i} \equiv \frac{\operatorname{cov}\left(m, R^{i}\right)}{\sigma(m) \sigma\left(R^{i}\right)}
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$$
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\rho_{m, R^i} \equiv \frac{\operatorname{cov}\left(m, R^{i}\right)}{\sigma(m) \sigma\left(R^{i}\right)}
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$$
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and where $\sigma$ denotes the standard deviation of the variable in parentheses
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Equation {eq}`eq:EMR5` implies
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$$E R^{i}=R^{f}-\rho_{m, R^i} \frac{\sigma(m)}{E(m)} \sigma\left(R^{i}\right)$$
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$$
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E R^{i}=R^{f}-\rho_{m, R^i} \frac{\sigma(m)}{E(m)} \sigma\left(R^{i}\right)
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$$
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Because $\rho_{m, R^i} \in [-1,1]$, it follows that $|\rho_{m, R^i}| \leq 1$ and that
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@@ -250,14 +270,18 @@ Evidently, points on the frontier correspond to gross returns that are perfectly
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We summarize this observation as
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$$\rho_{m, R^{i}}=\left\{\begin{array}{ll}
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$$
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\rho_{m, R^{i}}=\left\{\begin{array}{ll}
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+1 & \implies R^i \text { is on lower frontier } \\
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-1 & \implies R^i \text { is on an upper frontier }
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\end{array}\right.$$
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\end{array}\right.
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$$
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The image below illustrates a mean-variance frontier.
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<img src = "./AssetPricing_v1.jpg" style="zoom:60%">
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```{figure} _static/lecture_specific/asset_pricing_lph/AssetPricing_v1.jpg
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:scale: 60%
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```
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The mathematical structure of the mean-variance frontier described by inequality {eq}`eq:ERM6` implies
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that
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- for any mean-variance-efficient return $R^{m v}$ that is on the frontier but that is **not** $R^{f}$, there exists a **single-beta representation** for any return $R^i$ that takes the form:
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$$ E R^{i}=R^{f}+\beta_{i, R^{m v}}\left[E\left(R^{m v}\right)-R^{f}\right]
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$$
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E R^{i}=R^{f}+\beta_{i, R^{m v}}\left[E\left(R^{m v}\right)-R^{f}\right]
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$$ (eq:EMR7)
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- The special case of a single-beta representation {eq}`eq:EMR7` with $ R^{i}=R^{m v}$ is
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One is a **time-series regression** of gross return $R_t^i$ on multiple
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risk factors $f_t^j, j = a, b, \ldots $ that is designed to uncover exposures of return $R^i$ to each of a set of **risk-factors** $f_t^j, j = a, b, \ldots, $:
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- $$R_{t}^{i}=a_{i}+\beta_{i, a} f_{t}^{a}+\beta_{i, b} f_{t}^{b}+\ldots+\epsilon_{t}^{i}, \quad t=1,2, \ldots, T\\
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$$
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R_{t}^{i}=a_{i}+\beta_{i, a} f_{t}^{a}+\beta_{i, b} f_{t}^{b}+\ldots+\epsilon_{t}^{i}, \quad t=1,2, \ldots, T\\
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\epsilon_{t}^{i} \perp f_{t}^{j}, i=1,2, \ldots, I; j = a, b, \ldots
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$$
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$$
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\begin{aligned}
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E R^{i} & =\gamma+\beta_{i, a} \lambda_{a}+\beta_{i, b} \lambda_{b}+\cdots \\
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& \quad \text{for } i=1,2, \ldots, I \sim \text { returns } R^i \\
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E R^{i} & =\gamma+\beta_{i, a} \lambda_{a}+\beta_{i, b} \lambda_{b}+\cdots
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& \quad \text{for } i=1,2, \ldots, I \sim \text { returns } R^i
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\lambda_{j}, & j=a, b, \ldots \ldots = \text { price of exposure }
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\text { to risk factor } a, b, \ldots
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\end{aligned}
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$E R^i$ over some period and then estimate the **cross-section regression**
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$$\underbrace{E\left(R^{i}\right)}_{\text{average return over time series}}=\gamma+\underbrace{\beta_{i, a}}_{\text{regressor}\quad} \underbrace{\lambda_{a}}_{\text{regression}\\\text{coefficient}}+\underbrace{\beta_{i, b}}_{\text{regressor}\quad} \underbrace{\lambda_{b}}_{\text{regression}\\\text{coefficient}}+\cdots+\underbrace{\alpha_{i}}_{\text{pricing errors}}, i=1, \ldots, I; \quad \underbrace{\alpha_i \perp \beta_{i,j},j = a, b, \ldots}_{\text{least squares orthogonality condition}}$$
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$$
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\underbrace{E\left(R^{i}\right)}_{\text{average return over time series}}=\gamma+\underbrace{\beta_{i, a}}_{\text{regressor}\quad} \underbrace{\lambda_{a}}_{\text{regression}\text{coefficient}}+\underbrace{\beta_{i, b}}_{\text{regressor}\quad} \underbrace{\lambda_{b}}_{\text{regression}\text{coefficient}}+\cdots+\underbrace{\alpha_{i}}_{\text{pricing errors}}, i=1, \ldots, I; \quad \underbrace{\alpha_i \perp \beta_{i,j},j = a, b, \ldots}_{\text{least squares orthogonality condition}}
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$$
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- estimate $\gamma, \lambda_{a}, \lambda_{b}, \ldots$ by an appropriate regression technique, being thoughtful about recognizing that the regressors have been generated by a step 1 regression.
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