corrected some pages

Several small corrections were done to several proofs and definitions.

Author: JoramSoch

P/ind-self.md

@@ -36,7 +36,7 @@ username: "JoramSoch"
 **Theorem:** Let $E$ be a [random event](/D/reve). Then, $E$ is [independent of itself](/D/ind), if and only if its [probability](/D/prob) is zero or one:
 
 $$ \label{eq:ind-self}
-E \text{ self-independent} \quad \Leftrightarrow \quad p(E) = 0 \quad \text{or} \quad p(E) = 1 \; .
+E \; \text{self-independent} \quad \Leftrightarrow \quad p(E) = 0 \quad \text{or} \quad p(E) = 1 \; .
 $$
 
 

P/lognorm-mode.md

@@ -65,9 +65,9 @@ $$
 
 $$ \label{eq:lognorm-pdf-der2}
 \begin{split}
-f''_X(x) &= \frac{1}{\sqrt{2\pi}\sigma^2x^3} \mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \cdot \left( \ln x -\mu \right) \cdot \left(1 + \frac{ \ln x -\mu}{\sigma^2}\right) \\
-&+ \frac{\sqrt{2}}{\sqrt{\pi}x^3}\mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \cdot \left(1 + \frac{ \ln x -\mu}{\sigma^2}\right) \\
-&- \frac{1}{\sqrt{2\pi}\sigma^2x^3} \mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \; .
+f''_X(x) &= \frac{1}{\sqrt{2\pi}\sigma^2 x^3} \cdot \mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \cdot \left( \ln x -\mu \right) \cdot \left(1 + \frac{ \ln x -\mu}{\sigma^2}\right) \\
+&+ \frac{\sqrt{2}}{\sqrt{\pi}x^3} \cdot \mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \cdot \left(1 + \frac{ \ln x -\mu}{\sigma^2}\right) \\
+&- \frac{1}{\sqrt{2\pi}\sigma^2 x^3} \cdot \mathrm{exp} \left[ -\frac{\left( \ln x -\mu \right)^2}{2 \sigma^2} \right] \; .
 \end{split}
 $$
 
@@ -85,10 +85,10 @@ By plugging this value into the second derivative \eqref{eq:lognorm-pdf-der2},
 
 $$ \label{eq:lognorm-mode-s2}
 \begin{split}
-f''_X(e^{(\mu-\sigma^2)}) &= \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \cdot \left( \sigma^2 \right) \cdot \left(0\right) \\
-&+ \frac{\sqrt{2}}{\sqrt{\pi}(e^{(\mu-\sigma^2)})^3} \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \cdot \left(0 \right) \\
-&- \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \\
-&= - \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] < 0 \; ,
+f''_X(e^{(\mu-\sigma^2)}) &= \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \cdot \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \cdot \left( \sigma^2 \right) \cdot \left(0\right) \\
+&+ \frac{\sqrt{2}}{\sqrt{\pi}(e^{(\mu-\sigma^2)})^3} \cdot \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \cdot \left(0 \right) \\
+&- \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \cdot \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] \\
+&= - \frac{1}{\sqrt{2\pi}\sigma^2(e^{(\mu-\sigma^2)})^3} \cdot \mathrm{exp} \left[ -\frac{\sigma^2}{2} \right] < 0 \; ,
 \end{split}
 $$
 

P/med-mae.md

@@ -58,7 +58,7 @@ $$ \label{eq:med-mae-s1}
 E(\lvert X_i - a \rvert) = \int_{-\infty}^a (a - x) f(x) \, \mathrm{d}x + \int_{a}^\infty (x - a) f(x) \, \mathrm{d}x \; .
 $$
 
-Now note that $\lvert\frac{\partial}{\partial a}(a - x)f(x)\rvert = \lvert\frac{\partial}{\partial a}(x - a)f(x)\rvert = f(x)$. Consequently, $\int_{-\infty}^af(x) = P(X_i < a)$ and $\int_{a}^\infty f(x) = P(X_i > a)$, both of which must be finite by the [axioms of probability](/D/prob-ax). Therefore, these integrals meet the conditions for application of Leibniz's rule.
+Now note that $\lvert\frac{\partial}{\partial a}(a - x)f(x)\rvert = \lvert\frac{\partial}{\partial a}(x - a)f(x)\rvert = f(x)$. Consequently, $\int_{-\infty}^a f(x) = \mathrm{Pr}(X_i < a)$ and $\int_{a}^\infty f(x) = \mathrm{Pr}(X_i > a)$, both of which must be finite by the [axioms of probability](/D/prob-ax). Therefore, these integrals meet the conditions for application of Leibniz's rule.
 
 Applying Leibniz's integral rule, we can differentiate the objective function as follows:
 
@@ -74,17 +74,17 @@ Canceling non-integral terms and setting this derivative to 0, it must be true t
 $$\label{eq:dmed-da}
 \int_{-\infty}^a f(x) \, \mathrm{d}x - \int_{a}^\infty f(x) \, \mathrm{d}x = 0
 \quad \Rightarrow \quad
-P(X_i < a) = P(X_i > a) \; .
+\mathrm{Pr}(X_i < a) = \mathrm{Pr}(X_i > a) \; .
 $$
 
 Together with the [probability of the complement](/P/prob-comp), this yields the implication
 
 $$\label{eq:med-mae-qed}
-P(X_i < a) = P(X_i > a)
+\mathrm{Pr}(X_i < a) = \mathrm{Pr}(X_i > a)
 \quad \Rightarrow \quad 
-P(X_i < a) = 1 - P(X_i < a)
+\mathrm{Pr}(X_i < a) = 1 - \mathrm{Pr}(X_i < a)
 \quad \Rightarrow \quad
-P(X_i < a) = 0.5
+\mathrm{Pr}(X_i < a) = 0.5
 $$
 
 As a result, $a$ satisfies the [definition of a median](/D/med) at the critical point of the objective function.

P/norm-dent.md

@@ -61,7 +61,7 @@ $$ \label{eq:norm-dent-s1}
 \end{split}
 $$
 
-Note that $\mathrm{E}\left[ (x-\mu)^2 \right]$ [corresponds to the variance](/D/var) of $X$ and the [variance of the normal distribution](/P/norm-var) is $\sigma^2$. Thus, we can proceed:
+Note that $\mathrm{E}\left[ (x-\mu)^2 \right]$ [corresponds to the variance](/P/momcent-2nd) of $X$ and the [variance of the normal distribution](/P/norm-var) is $\sigma^2$. Thus, we can proceed:
 
 $$ \label{eq:norm-dent-s2}
 \begin{split}

P/snr-rsq.md

@@ -83,5 +83,5 @@ $$
 Rearranging this equation for the [coefficient of determination](/D/rsq), we have
 
 $$ \label{eq:R2-SNR-qed}
-R^2 = \frac{\mathrm{SNR}}{\mathrm{1+\mathrm{SNR}}} \; ,
+R^2 = \frac{\mathrm{SNR}}{\mathrm{1+\mathrm{SNR}}} \; .
 $$

P/ug-ttest1power.md

@@ -33,7 +33,7 @@ z_x = \Phi^{-1}(1 - x)
 \quad \Leftrightarrow \quad
 \Phi(z_x) = 1 - x
 \quad \Leftrightarrow \quad
-\mathbb{P}(Z \leq z_x) = 1 - x \Leftrightarrow \mathbb{P}(z_x \leq Z) = x \; .
+\mathrm{Pr}(Z \leq z_x) = 1 - x \Leftrightarrow \mathrm{Pr}(z_x \leq Z) = x \; .
 $$
 
 1) One-sided one-sample t-test:
@@ -97,20 +97,20 @@ $$
 So the [power](/D/power) of the test is
 
 $$ \label{eq:one-sided-power-prob}
-\mathbb{P}\left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} > z_\alpha\right) \; .
+\mathrm{Pr}\left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} > z_\alpha\right) \; .
 $$
 
 If the true mean is $\mu = \mu_0 + \delta$, where $\delta > 0$, then the power is
 
 $$ \label{eq:one-sided-power-shift}
-\mathbb{P}\left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{s} > z_\alpha\right) \; .
+\mathrm{Pr}\left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{s} > z_\alpha\right) \; .
 $$
 
 Now that we're subtracting the true mean, the first term in the sum has a [t-distribution](/D/t), which we approximate with a $\mathcal{N}(0,1)$ distribution, denoted as $Z$. We also approximate $s$ with $\sigma$, since it should be a good approximation, and the error here doesn't make much difference to the calculation. The power is therefore approximately
 
 $$ \label{eq:one-sided-power-final}
-  \mathbb{P}\left(Z > z_\alpha - \frac{\delta\sqrt{n}}{\sigma}\right)
-= \mathbb{P}\left(Z < \frac{\delta\sqrt{n}}{\sigma} - z_\alpha\right)
+  \mathrm{Pr}\left(Z > z_\alpha - \frac{\delta\sqrt{n}}{\sigma}\right)
+= \mathrm{Pr}\left(Z < \frac{\delta\sqrt{n}}{\sigma} - z_\alpha\right)
 = \Phi\left(\frac{\delta\sqrt{n}}{\sigma} - z_\alpha\right) \; .
 $$
 
@@ -151,13 +151,13 @@ $$
 So the [power](/D/power) of the test is
 
 $$ \label{eq:two-sided-power-prob}
-\mathbb{P}\left(\left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} > z_{\alpha/2}\right) \wedge \left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} < -z_{\alpha/2}\right)\right) \; .
+\mathrm{Pr}\left(\left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} > z_{\alpha/2}\right) \wedge \left(\frac{\bar{Y} - \mu_0}{s/\sqrt{n}} < -z_{\alpha/2}\right)\right) \; .
 $$
 
 If the true mean is $\mu = \mu_0 + \delta$ (where $\delta > 0$ without loss of generality), then the power is
 
 $$ \label{eq:two-sided-power-shift}
-\mathbb{P}\left(\left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{\sigma} > z_{\alpha/2}\right) \wedge \left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{\sigma} < -z_{\alpha/2}\right)\right) \; .
+\mathrm{Pr}\left(\left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{\sigma} > z_{\alpha/2}\right) \wedge \left(\frac{\bar{Y} - \mu_0 - \delta}{s/\sqrt{n}} + \frac{\delta\sqrt{n}}{\sigma} < -z_{\alpha/2}\right)\right) \; .
 $$
 
 As before, we use a $\mathcal{N}(0,1)$ distribution and we approximate $s$ with $\sigma$. The power is therefore approximately

P/wald-skew.md

@@ -21,7 +21,6 @@ username: "tomfaulkenberry"
 ---
 
 
-
 **Theorem:** Let $X$ be a [random variable](/D/rvar) following a [Wald distribution](/D/wald):
 
 $$ \label{eq:wald}
@@ -34,21 +33,20 @@ $$ \label{eq:wald-skew}
 \mathrm{Skew}(X) = \frac{3}{\sqrt{\alpha\gamma}} \; .
 $$
 
-**Proof:** 
 
-To compute the skewness of $X$, we [partition the skewness into expected values](/P/skew-mean):
+**Proof:** To compute the skewness of $X$, we [partition the skewness into expected values](/P/skew-mean):
 
 $$ \label{eq:skew-mean}
 \mathrm{Skew}(X) = \frac{\mathrm{E}(X^3)-3\mu\sigma^2-\mu^3}{\sigma^3} \; ,
 $$
 
-where $\mu$ and $\sigma$ are the mean and standard deviation of $X$, respectively. Since $X$ follows an [Wald distribution](/D/wald), the [mean](/P/wald-mean) of $X$ is given by 
+where $\mu$ and $\sigma$ are the mean and standard deviation of $X$, respectively. Since $X$ follows a [Wald distribution, the mean](/P/wald-mean) of $X$ is given by 
 
 $$ \label{eq:wald-mean}
 \mu = \mathrm{E}(X) = \frac{\alpha}{\gamma}
 $$
 
-and the [standard deviation](/P/wald-var) of $X$ is given by
+and [the standard deviation](/P/wald-var) of $X$ is given by
 
 $$ \label{eq:wald-var}
 \sigma = \sqrt{\mathrm{Var}(X)} = \sqrt{\frac{\alpha}{\gamma^3}}\; .
@@ -116,7 +114,7 @@ $$ \label{eq:wald-skew-split2}
 g(t) = \frac{\alpha}{(\gamma^2-2t)^{3/2}}\exp\left[\alpha\gamma-\sqrt{\alpha^2(\gamma^2-2t)}\right] \; .
 $$
 
-With this decomposition, $M_X'''(t) = f'(t) + g'(t)$. Applying the product rule to $f$ gives:
+With this decomposition, $M_X''\'(t) = f'(t) + g'(t)$. Applying the product rule to $f$ gives:
 
 $$ \label{eq:wald-skew-f}
 \begin{split}
@@ -161,4 +159,4 @@ $$ \label{eq:wald-skew-s5}
 \end{split}
 $$
 
-This completes the proof of \eqref{eq:wald-skew}.
+This completes the proof of \eqref{eq:wald-skew}.