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corrected some pages
Several small corrections were done to several proofs and definitions.
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P/bvn-mi.md

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@@ -58,7 +58,7 @@ $$ \label{eq:mvn-marg}
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X_1 \sim \mathcal{N}\left( \mu_1, \Sigma_{11} \right) \; ,
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$$
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such that the [marginals](/D/marg) of the [bivariate normal distribution](/D/bvn) are [univariate normal distribution](/D/norm):
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such that the [marginals](/D/dist-marg) of the [bivariate normal distribution](/D/bvn) are [univariate normal distributions](/D/norm):
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$$ \label{eq:bvn-marg}
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\left[ \begin{matrix} X \\ Y \end{matrix} \right] \sim

P/corr-range.md

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@@ -59,7 +59,7 @@ $$
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Using the [relationship between covariance and correlation](/P/cov-corr), we have:
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$$ \label{eq:var-XY-s2}
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\mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) = 1 + 1 + \pm 2 \, \mathrm{Corr}(X,Y) \; .
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\mathrm{Var}\left( \frac{X}{\sigma_X} \pm \frac{Y}{\sigma_Y} \right) = 1 + 1 \pm 2 \, \mathrm{Corr}(X,Y) \; .
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$$
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Thus, the combination of \eqref{eq:var-XY-0} with \eqref{eq:var-XY-s2} yields

P/exp-var.md

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@@ -77,8 +77,8 @@ Using the following anti-derivative
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$$ \label{eq:second-moment-s2}
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\begin{split}
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\int x^2 \cdot \exp(-\lambda x) \, \mathrm{d}x &= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int 2x \left( - \frac{1}{\lambda} x \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \\
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&= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left( \left[ \frac{1}{\lambda^2} 2x \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int 2 \left( \frac{1}{\lambda^2} \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \right) \\
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\int_{0}^{+\infty} x^2 \cdot \exp(-\lambda x) \, \mathrm{d}x &= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int_{0}^{+\infty} 2x \left( - \frac{1}{\lambda} x \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \\
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&= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left( \left[ \frac{1}{\lambda^2} 2x \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int_{0}^{+\infty} 2 \left( \frac{1}{\lambda^2} \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \right) \\
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&= \left[ - \frac{x^2}{\lambda} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left( \left[ \frac{2x}{\lambda^2} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left[ - \frac{2}{\lambda^3} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} \right) \\
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&= \left[ \left( - \frac{x^2}{\lambda} - \frac{2x}{\lambda^2} - \frac{2}{\lambda^3} \right) \exp(-\lambda x) \right]_{0}^{+\infty} \; ,
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\end{split}

P/glm-ols.md

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@@ -14,6 +14,12 @@ topic: "General linear model"
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theorem: "Ordinary least squares"
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sources:
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- authors: "Wikipedia"
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year: 2025
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title: "Ordinary least squares"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2025-12-19"
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url: "https://en.wikipedia.org/wiki/Ordinary_least_squares#Assumptions"
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proof_id: "P106"
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shortcut: "glm-ols"

P/mvn-mi.md

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@@ -58,7 +58,7 @@ $$ \label{eq:mvn-marg}
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X_1 \sim \mathcal{N}\left( \mu_1, \Sigma_{11} \right) \; ,
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$$
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such that the [marginals](/D/marg) of $X$ and $Y$ are:
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such that the [marginals](/D/dist-marg) of $X$ and $Y$ are:
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$$ \label{eq:X-Y-marg}
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X \sim \mathcal{N}\left( \mu_1, \Sigma_1 \right)

P/norm-var.md

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in: "StackExchange Mathematics"
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pages: "retrieved on 2020-01-09"
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url: "https://math.stackexchange.com/questions/518281/how-to-derive-the-mean-and-variance-of-a-gaussian-random-variable"
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- authors: "Wikipedia"
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year: 2025
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title: "Particular values of the gamma function"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2025-12-19"
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url: "https://en.wikipedia.org/wiki/Particular_values_of_the_gamma_function#Integers_and_half-integers"
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proof_id: "P18"
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shortcut: "norm-var"

P/rsq-der.md

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**Proof:** The [coefficient of determination](/D/rsq) $R^2$ is defined as the proportion of the variance explained by the independent variables, relative to the total variance in the data.
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<br>
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1) If we define the [explained sum of squares](/D/ess) as
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$$ \label{eq:ESS}
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R^2 = \frac{\mathrm{TSS}-\mathrm{RSS}}{\mathrm{TSS}} = 1 - \frac{\mathrm{RSS}}{\mathrm{TSS}} \; ,
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$$
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[because](/P/mlr-pss) $\mathrm{TSS} = \mathrm{ESS} + \mathrm{RSS}$.
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[because it holds that](/P/mlr-pss)
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$$ \label{eq:TSS-ESS-RSS}
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\mathrm{TSS} = \mathrm{ESS} + \mathrm{RSS} \; .
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$$
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<br>
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2) Using \eqref{eq:SS}, the coefficient of determination can be also written as:
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$$ \label{eq:R2'}
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R^2_{\mathrm{adj}} = 1 - \frac{\frac{1}{n-p} \sum_{i=1}^{n} (y_i - \hat{y}_i)^2}{\frac{1}{n-1} \sum_{i=1}^{n} (y_i - \bar{y})^2} = 1 - \frac{\mathrm{RSS}/\mathrm{df}_r}{\mathrm{TSS}/\mathrm{df}_t}
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$$
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where $\mathrm{df}_r = n-p$ and $\mathrm{df}_t = n-1$ are the residual and total [degrees of freedom](/D/dof).
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<br>
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This gives the adjusted $R^2$ which adjusts $R^2$ for the number of explanatory variables.
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where $\mathrm{df}_r = n-p$ and $\mathrm{df}_t = n-1$ are the residual and total [degrees of freedom](/D/dof). This gives the adjusted $R^2$ which adjusts $R^2$ for the [number of explanatory variables](/D/mlr).

P/slr-rescorr.md

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---
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**Theorem:** In [simple linear regression](/D/slr), the [residuals](/D/rss) and the [covariate](/D/slr) are [uncorrelated](/D/corr) when estimated using [ordinary least squares](/P/slr-ols).
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**Theorem:** In [simple linear regression](/D/slr), the [residuals](/D/rss) and the [covariate](/D/slr) are [uncorrelated](/D/corr) when [estimated using ordinary least squares](/P/slr-ols).
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**Proof:** The residuals are defined as the estimated [error terms](/D/slr)
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$$ \label{eq:slr-res}
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\hat{\varepsilon}_i = y_i - \hat{\beta}_0 - \hat{\beta}_1 x_i
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$$
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where $\hat{\beta}_0$ and $\hat{\beta}_1$ are parameter estimates obtained using [ordinary least squares](/P/slr-ols):
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where $\hat{\beta}_0$ and $\hat{\beta}_1$ are parameter estimates [obtained using ordinary least squares](/P/slr-ols):
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$$ \label{eq:slr-ols}
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\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x} \quad \text{and} \quad \hat{\beta}_1 = \frac{s_{xy}}{s_x^2} \; .
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\end{split}
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$$
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Because an inner product of zero also implies zero [correlation](/D/corr), this demonstrates that [residuals](/D/rss) and [covariate](/D/slr) values are uncorrelated under [ordinary least squares](/P/slr-ols).
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Because an inner product of zero also implies zero [correlation](/D/corr), this demonstrates that [residuals](/D/rss) and [covariate](/D/slr) values are uncorrelated [under ordinary least squares](/P/slr-ols).

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