corrected some pages

Several small mistakes/errors were corrected in several proofs/definitions.

Author: JoramSoch

D/chi2.md

@@ -45,7 +45,7 @@ $$\label{eq:chi2}
 Y = \sum_{i=1}^{k} X_{i}^{2} \sim \chi^{2}(k) \quad \text{where} \quad k > 0 \; .
 $$
 
-A [random variables](/D/rvar) $Y$ is said said to follow a chi-square distribution with $k$ number of degress of freedom, if and only if its [probability density function](/D/pdf) is given by
+A [random variable](/D/rvar) $Y$ is said said to follow a chi-square distribution with $k$ number of degress of freedom, if and only if its [probability density function](/D/pdf) is given by
 
 $$ \label{eq:chi2-pdf}
 \chi^{2}(x; k) = \frac{1}{2^{k/2}\Gamma (k/2)} \, x^{k/2-1} \, e^{-x/2}

D/sgam.md

@@ -21,6 +21,13 @@ sources:
     pages: "retrieved on 2020-05-26"
     url: "https://github.com/JoramSoch/MACS/blob/master/MD_gamrnd.m"
     doi: "10.5281/zenodo.845404"
+  - authors: "NIST/SEMATECH"
+    year: 2012
+    title: "Gamma distribution"
+    in: "e-Handbook of Statistical Methods"
+    pages: "ch. 1.3.6.6.11"
+    url: "https://www.itl.nist.gov/div898/handbook/eda/section3/eda366b.htm"
+    doi: "10.18434/M32189"
 
 def_id: "D64"
 shortcut: "sgam"

P/cuni-mean.md

@@ -37,7 +37,7 @@ $$
 **Proof:** The [expected value](/D/mean) is the probability-weighted average over all possible values:
 
 $$ \label{eq:mean}
-\mathrm{E}(X) = \int_{\mathbb{R}} x \cdot f_X(x) \, \mathrm{d}x \; .
+\mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .
 $$
 
 With the [probability density function of the continuous uniform distribution](/P/cuni-pdf), this becomes:

P/exp-mean.md

@@ -44,7 +44,7 @@ $$
 **Proof:** The [expected value](/D/mean) is the probability-weighted average over all possible values:
 
 $$ \label{eq:mean}
-\mathrm{E}(X) = \int_{\mathbb{R}} x \cdot f_\mathrm{X}(x) \, \mathrm{d}x \; .
+\mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_\mathrm{X}(x) \, \mathrm{d}x \; .
 $$
 
 With the [probability density function of the exponential distribution](/P/exp-pdf), this reads:

P/gam-logmean.md

@@ -36,9 +36,11 @@ $$
 Then, the [expectation](/D/mean) of the natural logarithm of $X$ is
 
 $$ \label{eq:gam-logmean}
-\mathrm{E}(\ln X) = \psi(a) - \ln(b) \; .
+\mathrm{E}(\ln X) = \psi(a) - \ln(b)
 $$
 
+where $\psi(x)$ is the digamma function.
+
 
 **Proof:** Let $Y = \ln(X)$, such that $\mathrm{E}(Y) = \mathrm{E}(\ln X)$ and consider the special case that $b = 1$. In this case, the [probability density function of the gamma distribution](/P/gam-pdf) is
 

P/gam-mean.md

@@ -43,7 +43,7 @@ $$
 **Proof:** The [expected value](/D/mean) is the probability-weighted average over all possible values:
 
 $$ \label{eq:mean}
-\mathrm{E}(X) = \int_{\mathbb{R}} x \cdot f_X(x) \, \mathrm{d}x \; .
+\mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .
 $$
 
 With the [probability density function of the gamma distribution](/P/gam-pdf), this reads:

P/gam-sgam.md

@@ -27,17 +27,17 @@ $$ \label{eq:X-gam}
 X \sim \mathrm{Gam}(a,b) \; .
 $$
 
-Then, the quantity $Z = b X$ will have a [standard gamma distribution](/D/sgam) with shape $a$ and rate $1$:
+Then, the quantity $Y = b X$ will have a [standard gamma distribution](/D/sgam) with shape $a$ and rate $1$:
 
-$$ \label{eq:Z-snorm}
-Z = b X \sim \mathrm{Gam}(a,1) \; .
+$$ \label{eq:Y-snorm}
+Y = b X \sim \mathrm{Gam}(a,1) \; .
 $$
 
 
-**Proof:** Rearranging to get $X$ in terms of $Z$, we have
+**Proof:** Rearranging to get $X$ in terms of $Y$, we have
 
-$$ \label{eq:X-Z}
-X = \frac{1}{b} Z \; .
+$$ \label{eq:X-Y}
+X = \frac{1}{b} Y \; .
 $$
 
 The [cumulative distribution function of the gamma-distributed](/P/gam-cdf) $X$ is
@@ -46,13 +46,13 @@ $$ \label{eq:gam-cdf}
 F_X(t) = \int_{-\infty}^{t} \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \, \mathrm{d}x \; .
 $$
 
-Substituting \eqref{eq:X-Z} into \eqref{eq:gam-cdf}, we obtain
+Substituting \eqref{eq:X-Y} into \eqref{eq:gam-cdf}, we obtain
 
 $$ \label{eq:sgam-cdf}
 \begin{split}
-F_Z(t) &= \int_{-\infty}^{t} \frac{b^a}{\Gamma(a)} \left(\frac{1}{b} z\right)^{a-1} \exp\left[-b \left(\frac{1}{b} z\right)\right] \, \mathrm{d}\left(\frac{1}{b} z\right) \\
-&= \int_{-\infty}^{t} \frac{b^a}{b} \left(\frac{1}{b}\right)^{a-1} \cdot \frac{1}{\Gamma(a)} z^{a-1} \exp[-z] \, \mathrm{d}z \\
-&= \int_{-\infty}^{t} \frac{1}{\Gamma(a)} z^{a-1} \exp[-z] \, \mathrm{d}z
+F_Z(t) &= \int_{-\infty}^{t} \frac{b^a}{\Gamma(a)} \left(\frac{1}{b} y\right)^{a-1} \exp\left[-b \left(\frac{1}{b} y\right)\right] \, \mathrm{d}\left(\frac{1}{b} y\right) \\
+&= \int_{-\infty}^{t} \frac{b^a}{b} \left(\frac{1}{b}\right)^{a-1} \cdot \frac{1}{\Gamma(a)} y^{a-1} \exp[-y] \, \mathrm{d}y \\
+&= \int_{-\infty}^{t} \frac{1}{\Gamma(a)} y^{a-1} \exp[-y] \, \mathrm{d}y
 \end{split}
 $$
 

P/gam-xlogx.md

@@ -40,7 +40,7 @@ $$ \label{eq:gam-xlogx}
 $$
 
 
-**Proof:** With the [definition of the expected value](/D/mean) and the [probability density function of the gamma distribution](/P/gam-pdf), we have:
+**Proof:** With the [definition of the expected value](/D/mean), the [law of the unconscious statistician](/P/mean-lotus) and the [probability density function of the gamma distribution](/P/gam-pdf), we have:
 
 $$ \label{eq:gam-xlogx-s1}
 \begin{split}
@@ -68,7 +68,7 @@ $$ \label{eq:gam-fct}
 \Gamma(x+1) = \Gamma(x) \cdot x \quad \Leftrightarrow \quad \frac{\Gamma(x+1)}{\Gamma(x)} = x \; ,
 $$
 
-the expression in equation \eqref{eq:gam-mean-s1} develops into:
+the expression in equation \eqref{eq:gam-xlogx-s1} develops into:
 
 $$ \label{eq:gam-xlogx-qed}
 \mathrm{E}(X \ln X) = \frac{a}{b} \left[ \psi(a) - \ln(b) \right] \; .

P/norm-mean.md

@@ -43,7 +43,7 @@ $$
 **Proof:** The [expected value](/D/mean) is the probability-weighted average over all possible values:
 
 $$ \label{eq:mean}
-\mathrm{E}(X) = \int_{\mathbb{R}} x \cdot f_X(x) \, \mathrm{d}x \; .
+\mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .
 $$
 
 With the [probability density function of the normal distribution](/P/norm-pdf), this reads: