From 8f19a52025af702febd18dab007803b921d0a701 Mon Sep 17 00:00:00 2001 From: majapavlo Date: Thu, 6 Oct 2022 22:18:28 +0100 Subject: [PATCH 1/2] fix indentation and eqreference --- P/lognorm-mean.md | 12 ++++++------ P/lognorm-var.md | 22 +++++++++++----------- 2 files changed, 17 insertions(+), 17 deletions(-) diff --git a/P/lognorm-mean.md b/P/lognorm-mean.md index 8bbbb65c..223e2882 100644 --- a/P/lognorm-mean.md +++ b/P/lognorm-mean.md @@ -49,8 +49,8 @@ With the [probability density function of the log-normal distribution](/P/lognor $$ \label{eq:lognorm-mean-s1} \begin{split} -\mathrm{E}(X) = \int_{0}^{+\infty} x \cdot \frac{1}{x\sqrt{2 \pi \sigma^2} } \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \\ -&= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{0}^{+\infty} \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x +\mathrm{E}(X) &= \int_{0}^{+\infty} x \cdot \frac{1}{x\sqrt{2 \pi \sigma^2} } \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \\ +&= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{0}^{+\infty} \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \end{split} $$ @@ -58,10 +58,10 @@ Substituting $z = \frac{\ln x -\mu}{\sigma}$, i.e. $x = \exp \left( \mu + \sigma $$ \label{eq:lognorm-mean-s2} \begin{split} -\mathrm{E}(X) = \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ +\mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 \right) \sigma \exp \left( \mu +\sigma z \right) \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 + \sigma z + \mu \right) \mathrm{d}z \\ -&= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z - 2 \mu \right) \right] \mathrm{d}z \\ +&= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z - 2 \mu \right) \right] \mathrm{d}z \end{split} $$ @@ -69,9 +69,9 @@ Now multiplying $\exp \left( \frac{1}{2} \sigma^2 \right)$ and $\exp \left( -\fr $$ \label{eq:lognorm-mean-s3} \begin{split} -\mathrm{E}(X) = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z + \sigma^2 - 2 \mu - \sigma^2 \right) \right] \mathrm{d}z \\ +\mathrm{E}(X) &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2 \sigma z + \sigma^2 - 2 \mu - \sigma^2 \right) \right] \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 2\sigma z + \sigma^2 \right) \right] \exp \left( \mu + \frac{1}{2} \sigma^2 \right) \mathrm{d}z \\ -&= \exp \left( \mu + \frac{1}{2} \sigma^2 \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - \sigma \right)^2 \right] \mathrm{d}z \\ +&= \exp \left( \mu + \frac{1}{2} \sigma^2 \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - \sigma \right)^2 \right] \mathrm{d}z \end{split} $$ diff --git a/P/lognorm-var.md b/P/lognorm-var.md index 84d2ebc5..1cbe3654 100644 --- a/P/lognorm-var.md +++ b/P/lognorm-var.md @@ -69,7 +69,7 @@ The second moment $\mathrm{E}[X^2]$ can be derived as follows: $$ \label{eq:second-moment} \begin{split} -\mathrm{E} [X^2] = \int_{- \infty}^{+\infty} x^2 \cdot f_\mathrm{X}(x) \, \mathrm{d}x \\ +\mathrm{E} [X^2] &= \int_{- \infty}^{+\infty} x^2 \cdot f_\mathrm{X}(x) \, \mathrm{d}x \\ &= \int_{0}^{+\infty} x^2 \cdot \frac{1}{x\sqrt{2 \pi \sigma^2} } \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right] \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{0}^{+\infty} x \cdot \exp \left[ -\frac{1}{2} \frac{\left(\ln x-\mu\right)^2}{\sigma^2} \right]\mathrm{d}x \end{split} @@ -77,25 +77,25 @@ $$ Substituting $z = \frac{\ln x -\mu}{\sigma}$, i.e. $x = \exp \left( \mu + \sigma z \right )$, we have: -$$ \label{eq:second-moment-s2} +$$ \label{eq:second-moment-2} \begin{split} -\mathrm{E} [X^2] = \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( \mu +\sigma z \right) \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ +\mathrm{E} [X^2] &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{(-\infty -\mu )/ (\sigma)}^{(\ln x -\mu )/ (\sigma)} \exp \left( \mu +\sigma z \right) \exp \left( -\frac{1}{2} z^2 \right) \mathrm{d} \left[ \exp \left( \mu +\sigma z \right) \right] \\ &= \frac{1}{\sqrt{2 \pi \sigma^2} } \int_{-\infty}^{+\infty} \exp \left( -\frac{1}{2} z^2 \right) \sigma \exp \left( 2\mu + 2 \sigma z \right) \mathrm{d}z \\ -&= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z - 4 \mu \right) \right] \mathrm{d}z \\ +&= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z - 4 \mu \right) \right] \mathrm{d}z \end{split} $$ Now multiplying by $\exp \left( 2 \sigma^2 \right)$ and $\exp \left(- 2 \sigma^2 \right)$, this becomes: -$$ \label{eq:second-moment-s3} +$$ \label{eq:second-moment-3} \begin{split} -\mathrm{E} [X^2] = \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z + 4 \sigma^2 -4 \sigma^2 - 4 \mu \right) \right] \mathrm{d}z \\ +\mathrm{E} [X^2] &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4 \sigma z + 4 \sigma^2 -4 \sigma^2 - 4 \mu \right) \right] \mathrm{d}z \\ &= \frac{1}{\sqrt{2 \pi} } \int_{-\infty}^{+\infty} \exp \left[ -\frac{1}{2} \left( z^2 - 4\sigma z + 4\sigma^2 \right) \right] \exp \left( 2 \sigma^2 +2 \mu \right) \mathrm{d}z \\ -&= \exp \left( 2 \sigma^2 +2 \mu \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - 2 \sigma \right)^2 \right] \mathrm{d}z \\ +&= \exp \left( 2 \sigma^2 +2 \mu \right) \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi} } \exp \left[ -\frac{1}{2} \left( z - 2 \sigma \right)^2 \right] \mathrm{d}z \end{split} $$ -The [probability density function of a normal distribution](/P/norm-pdf) is +The [probability density function of a normal distribution](/P/norm-pdf) is $$ f_X(x) = \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] @@ -109,11 +109,11 @@ $$ Using the definition of the [probability density function](/D/pdf) -$$ +$$ \label{eq:def-pdf} \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} \left({x - 2 \sigma} \right)^2 \right] \mathrm{d}x = 1 $$ -and applying \eqref{eq:def-pdf} to \eqref{second-moment-s3}, we have: +and applying \eqref{eq:def-pdf} to \eqref{second-moment-3}, we have: $$ \label{eq:second-moment-4} \mathrm{E}[X]^2 = \exp \left( 2 \sigma^2 +2 \mu \right) \; . @@ -123,7 +123,7 @@ Applying \eqref{eq:second-moment-4} and \eqref{eq:lognorm-mean-ref} to \eqref{eq $$ \label{eq:lognorm-var-2} \begin{split} -\mathrm{Var}(X) = \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \\ +\mathrm{Var}(X) &= \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \\ &= \exp \left(2\sigma^2 + 2\mu \right) - \left[ \exp \left(\mu + \frac{1}{2} \sigma^2 \right) \right]^2 \\ &= \exp \left(2\sigma^2 + 2\mu \right) - \exp \left(2\mu + \sigma^2\right) \; . \end{split} From 03c8738a6c090b5f2e4707e9cecb9ceac95d9ec0 Mon Sep 17 00:00:00 2001 From: majapavlo Date: Fri, 7 Oct 2022 07:56:06 +0100 Subject: [PATCH 2/2] fix equation reference --- P/lognorm-var.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/P/lognorm-var.md b/P/lognorm-var.md index 1cbe3654..b49bb172 100644 --- a/P/lognorm-var.md +++ b/P/lognorm-var.md @@ -113,7 +113,7 @@ $$ \label{eq:def-pdf} \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} \left({x - 2 \sigma} \right)^2 \right] \mathrm{d}x = 1 $$ -and applying \eqref{eq:def-pdf} to \eqref{second-moment-3}, we have: +and applying \eqref{eq:def-pdf} to \eqref{eq:second-moment-3}, we have: $$ \label{eq:second-moment-4} \mathrm{E}[X]^2 = \exp \left( 2 \sigma^2 +2 \mu \right) \; .