replaced function notation

In all definitions and proofs on the "chi-squared distribution", equation parts like "e^x" are replaced by "\exp[x]" or "\exp \left[ x \right]" throughout.

Author: JoramSoch

D/chi2.md

@@ -48,7 +48,7 @@ $$
 The [probability density function of the chi-squared distribution](/P/chi2-pdf) with $k$ degrees of freedom is
 
 $$ \label{eq:chi2-pdf}
-\chi^{2}(x; k) = \frac{1}{2^{k/2}\Gamma (k/2)} \, x^{k/2-1} \, e^{-x/2}
+\chi^{2}(x; k) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, x^{k/2-1} \, \exp \left[ -x/2 \right]
 $$
 
-where $k > 0$ and the density is zero if $x \leq 0$.
+where $k > 0$ and the density is zero, if $x \leq 0$.

P/chi2-gam.md

@@ -21,7 +21,7 @@ username: "kjpetrykowski"
 ---
 
 
-**Theorem:** The [chi-squared distribution](/D/chi2) with $k$ degrees of freedom is a special case of the [gamma distribution](/D/gam) with shape $\frac{k}{2}$ and rate $\frac{1}{2}$:
+**Theorem:** The [chi-squared distribution](/D/chi2) with $k$ degrees of freedom is a special case of the [gamma distribution](/D/gam) with shape $k/2$ and rate $1/2$:
 
 $$ \label{eq:chi2-gam}
 X \sim \mathrm{Gam}\left( \frac{k}{2}, \frac{1}{2} \right) \quad \Rightarrow \quad X \sim \chi^{2}(k) \; .
@@ -31,13 +31,13 @@ $$
 **Proof:** The [probability density function of the gamma distribution](/P/gam-pdf) for $x > 0$, where $\alpha$ is the shape parameter and $\beta$ is the rate parameter, is as follows:
 
 $$ \label{eq:gam-pdf}
-\mathrm{Gam}(x; \alpha, \beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \, x^{\alpha-1} \, e^{-\beta x}
+\mathrm{Gam}(x; \alpha, \beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \, x^{\alpha-1} \, \exp \left[ -\beta x \right]
 $$
 
 If we let $\alpha = k/2$ and $\beta = 1/2$, we obtain
 
 $$ \label{eq:gam-pdf-chi2}
-\mathrm{Gam}\left(x; \frac{k}{2}, \frac{1}{2}\right) = \frac{x^{k/2-1} \, e^{-x/2}}{\Gamma(k/2) 2^{k/2}} = \frac{1}{2^{k/2} \Gamma(k/2)} \, x^{k/2-1} \, e^{-x/2}
+\mathrm{Gam}\left(x; \frac{k}{2}, \frac{1}{2}\right) = \frac{x^{k/2-1} \, \exp \left[ -x/2 \right]}{\Gamma(k/2) \, 2^{k/2}} = \frac{1}{2^{k/2} \, \Gamma(k/2)} \, x^{k/2-1} \, \exp \left[ -x/2 \right]
 $$
 
 which is equivalent to the [probability density function of the chi-squared distribution](/P/chi2-pdf).

P/chi2-momraw.md

@@ -36,33 +36,33 @@ $$
 Then, if $m > -k/2$, the [raw moment](/D/mom-raw) $\mathrm{E}(X^{m})$ exists and is equal to:
 
 $$ \label{eq:chi2-momraw}
-\mathrm{E}(X^{m}) = 2^m \frac{\Gamma\left( \frac{k}{2}+m \right)}{\Gamma\left( \frac{k}{2} \right)} \; .
+\mathrm{E}(X^{m}) = 2^m \, \frac{\Gamma\left( \frac{k}{2}+m \right)}{\Gamma\left( \frac{k}{2} \right)} \; .
 $$
 
 
 **Proof:** Combining the [definition of the raw moment](/D/mom-raw) with the [probability density function of the chi-squared distribution](/P/chi2-pdf), we have:
 
 $$ \label{eq:chi2-momraw-int}
 \begin{split}
-\mathrm{E}(X^{m}) &= \int_{0}^{\infty} x^m \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \, x^{k/2-1} \, e^{-x/2} \, \mathrm{d}x \\
-&= \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} x^{(k/2)+m-1} \, e^{-x/2} \, \mathrm{d}x \; .
+\mathrm{E}(X^{m}) &= \int_{0}^{\infty} x^m \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \, x^{k/2-1} \, \exp \left[ -x/2 \right] \, \mathrm{d}x \\
+&= \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} x^{(k/2)+m-1} \, \exp \left[ -x/2 \right] \, \mathrm{d}x \; .
 \end{split}
 $$
 
 Now, we substitute $u = x/2$, such that $x = 2u$. As a result, we obtain:
 
 $$ \label{eq:chi2-momraw-int-u}
 \begin{split}
-\mathrm{E}(X^{m}) &= \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} 2^{(k/2)+m-1} \, u^{(k/2)+m-1} \, e^{-u} \, \mathrm{d}(2u) \\
-&= \frac{2^{(k/2)+m}}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} u^{(k/2)+m-1} \, e^{-u} \, \mathrm{d}u \\
-&= \frac{2^m}{\Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} u^{(k/2)+m-1} \, e^{-u} \, \mathrm{d}u \; .
+\mathrm{E}(X^{m}) &= \frac{1}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} 2^{(k/2)+m-1} \, u^{(k/2)+m-1} \, \exp[-u] \, \mathrm{d}(2u) \\
+&= \frac{2^{(k/2)+m}}{2^{k/2} \Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} u^{(k/2)+m-1} \, \exp[-u] \, \mathrm{d}u \\
+&= \frac{2^m}{\Gamma\left( \frac{k}{2} \right)} \int_{0}^{\infty} u^{(k/2)+m-1} \, \exp[-u] \, \mathrm{d}u \; .
 \end{split}
 $$
 
 With the definition of the gamma function as
 
 $$ \label{eq:gam-fct}
-\Gamma(x) = \int_{0}^{\infty} t^{x-1} \, e^{-t} \, \mathrm{d}t, \; z > 0 \; ,
+\Gamma(x) = \int_{0}^{\infty} t^{x-1} \, e^{-t} \, \mathrm{d}t, \; x > 0 \; ,
 $$
 
 this leads to the desired result when $m > -k/2$. Observe that, if $m$ is a non-negative integer, then $m > -k/2$ is always true. Therefore, all [moments](/D/mom) of a [chi-squared distribution](/D/chi2) exist and the $m$-th raw moment is given by the equation above.

P/chi2-pdf.md

@@ -42,7 +42,7 @@ $$
 Then, the [probability density function](/D/pdf) of $Y$ is
 
 $$ \label{eq:chi2-pdf}
-f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .
+f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, \exp \left[ -y/2 \right] \; .
 $$
 
 
@@ -112,5 +112,5 @@ $$
 From this, we get the final result in \eqref{eq:chi2-pdf}:
 
 $$ \label{eq:y-cdf-s5}
-f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .
-$$
+f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, \exp \left[ -y/2 \right] \; .
+$$