added 2 proofs

I/Table_of_Contents.md

@@ -78,8 +78,10 @@ Proofs by **[Number](/I/Proof_by_Number.html)** and **[Topic](/I/Proof_by_Topic.
    &emsp;&ensp; 3.4.1. *[Definition](/D/exp.html)* <br>
    &emsp;&ensp; 3.4.2. **[Probability density function](/P/exp-pdf.html)** <br>
    &emsp;&ensp; 3.4.3. **[Cumulative distribution function](/P/exp-cdf.html)** <br>
-   &emsp;&ensp; 3.4.4. **[Mean](/P/exp-mean.html)** <br>
-   &emsp;&ensp; 3.4.5. **[Median](/P/exp-med.html)** <br>
+   &emsp;&ensp; 3.4.4. **[Quantile function](/P/exp-qf.html)** <br>
+   &emsp;&ensp; 3.4.5. **[Mean](/P/exp-mean.html)** <br>
+   &emsp;&ensp; 3.4.6. **[Median](/P/exp-med.html)** <br>
+   &emsp;&ensp; 3.4.7. **[Mode](/P/exp-mode.html)** <br>
 
 4. Multivariate continuous distributions
 

P/exp-med.md

@@ -27,41 +27,33 @@ $$ \label{eq:exp}
 X \sim \mathrm{Exp}(\lambda) \; .
 $$
 
-Then, the mean or expected value of $X$ is
+Then, the median of $X$ is
 
-$$ \label{eq:exp-mean}
-\mathrm{E}(X) = \frac{1}{\lambda} \; .
+$$ \label{eq:exp-med}
+\mathrm{median}(X) = \frac{\ln 2}{\lambda} \; .
 $$
 
 
-**Proof:** The [expected value](/D/ev.html) is the probability-weighted average over all possible values:
+**Proof:** The [median](/D/med.html) is the value at which the cumulative distribution function is $1/2$:
 
-$$ \label{eq:mean}
-\mathrm{E}(X) = \int_{\mathbb{R}} x \cdot f_\mathrm{X}(x) \, \mathrm{d}x \; .
+$$ \label{eq:median}
+F_X(\mathrm{median}(X)) = \frac{1}{2} \; .
 $$
 
-With the [probability density function of the exponential distribution](/P/exp-pdf.html), this reads:
+The [cumulative distribution function of the exponential distribution](/D/exp-cdf.html) is
 
-$$ \label{eq:exp-mean-s1}
-\begin{split}
-\mathrm{E}(X) &= \int_{0}^{+\infty} x \cdot \lambda \exp(-\lambda x) \, \mathrm{d}x \\
-&= \lambda \int_{0}^{+\infty} x \cdot \exp(-\lambda x) \, \mathrm{d}x \; .
-\end{split}
+$$ \label{eq:exp-cdf}
+F_X(x) = 1 - \exp[-\lambda x], \quad x \geq 0 \; .
 $$
 
-Using the following anti-deriative
+Thus, the inverse CDF is
 
-$$ \label{eq:exp-mean-s2}
-\int x \cdot \exp(-\lambda x) \, \mathrm{d}x = \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) \; ,
+$$ \label{eq:exp-cdf-inv}
+x = -\frac{\ln(1-p)}{\lambda}
 $$
 
-the expected value becomes
+and setting $p = 1/2$, we obtain:
 
-$$ \label{eq:exp-mean-s3}
-\begin{split}
-\mathrm{E}(X) &= \lambda \left[ \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) \right]_{0}^{+\infty} \\
-&= \lambda \left[ \lim_{x \to \infty} \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) - \left( - \frac{1}{\lambda} \cdot 0 - \frac{1}{\lambda^2} \right) \exp(-\lambda \cdot 0) \right] \\
-&= \lambda \left[ 0 + \frac{1}{\lambda^2} \right] \\
-&= \frac{1}{\lambda} \; .
-\end{split}
+$$ \label{eq:exp-med-qed}
+\mathrm{median}(X) = -\frac{\ln(1-\frac{1}{2})}{\lambda} = \frac{\ln 2}{\lambda} \; .
 $$

P/exp-mode.md

@@ -0,0 +1,70 @@
+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-02-12 15:53:00
+
+title: "Mode of the exponential distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Exponential distribution"
+theorem: "Mode"
+
+sources:
+
+proof_id: "P51"
+shortcut: "exp-mode"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a random variable following an [exponential distribution](/D/exp.html):
+
+$$ \label{eq:exp}
+X \sim \mathrm{Exp}(\lambda) \; .
+$$
+
+Then, the [mode](/D/mode.html) of $X$ is
+
+$$ \label{eq:exp-mode}
+\mathrm{mode}(X) = 0 \; .
+$$
+
+
+**Proof:**  The [mode](/D/mode.html) is the value which maximizes the probability density function:
+
+$$ \label{eq:mode}
+\mathrm{mode}(X) = \operatorname*{arg\,max}_x f_X(x) \; .
+$$
+
+The [probability density function of the exponential distribution](/P/exp-pdf.html) is:
+
+$$ \label{eq:exp-pdf}
+f_X(x) = \left\{
+\begin{array}{rl}
+0 \; , & \text{if} \; x < 0 \\
+\lambda \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; .
+\end{array}
+\right.
+$$
+
+Since
+
+$$ \label{eq:exp-pdf-eq0}
+\lim_{x \to 0} f_X(x) = \infty
+$$
+
+and
+
+$$ \label{eq:exp-pdf-neq0}
+f_X(x) < \infty \quad \text{for any} \quad x \neq 0 \; ,
+$$
+
+it follows that
+
+$$ \label{eq:exp-mode-qed}
+\mathrm{mode}(X) = 0 \; .
+$$

P/exp-qf.md

@@ -0,0 +1,63 @@
+---
+layout: proof
+mathjax: true
+
+author: "Joram Soch"
+affiliation: "BCCN Berlin"
+e_mail: "joram.soch@bccn-berlin.de"
+date: 2020-02-12 15:48:00
+
+title: "Quantile function of the exponential distribution"
+chapter: "Probability Distributions"
+section: "Univariate continuous distributions"
+topic: "Exponential distribution"
+theorem: "Quantile function"
+
+sources:
+
+proof_id: "P50"
+shortcut: "exp-qf"
+username: "JoramSoch"
+---
+
+
+**Theorem:** Let $X$ be a random variable following an [exponential distribution](/D/exp.html):
+
+$$ \label{eq:exp}
+X \sim \mathrm{Exp}(\lambda) \; .
+$$
+
+Then, the [quantile function](/D/qf.html) of $X$ is
+
+$$ \label{eq:exp-qf}
+Q_X(p) = -\frac{\ln(1-p)}{\lambda} \; .
+$$
+
+
+**Proof:** The [cumulative distribution function of the exponential distribution](/P/exp-cdf.html) is:
+
+$$ \label{eq:exp-cdf}
+F_X(x) = \left\{
+\begin{array}{rl}
+0 \; , & \text{if} \; x < 0 \\
+1 - \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; .
+\end{array}
+\right.
+$$
+
+Thus, the [quantile function](/D/qf.html) is:
+
+$$ \label{eq:exp-qf-s1}
+Q_X(p) = F_X^{-1}(x) \; .
+$$
+
+This can be derived by rearranging equation \eqref{eq:exp-cdf}:
+
+$$ \label{eq:exp-qf-s2}
+\begin{split}
+p &= 1 - \exp[-\lambda x] \\
+\exp[-\lambda x] &= 1-p \\
+-\lambda x &= \ln(1-p) \\
+x &= -\frac{\ln(1-p)}{\lambda} \; .
+\end{split}
+$$