\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Notice that \(x_m\) is such that \[ \Prob (X \leq x_m) = F(x_m) = \frac {1}{2}. \]
\(y_m\) is such that \[ \Prob (Y \leq y_m) = \Prob (e^X \leq y_m) = \Prob (X \leq \ln y_m) = F(\ln y_m) = \frac {1}{2}. \]
Therefore, \[ F(x_m) = F(\ln y_m) = \frac {1}{2}. \]
Therefore, \(x_m = \ln y_m\), and \(y_m = e^{x_m}\).
Notice that the cumulative distribution function \(G(y)\) of \(Y\) satisfies that \[ G(y) = \Prob (Y \leq y) = \Prob (e^X \leq y) = \Prob (X \leq \ln y) = F(\ln y). \]
Therefore, differentiating both sides w.r.t. \(y\) gives that the probability density function of \(Y\), \(g(y)\) satisfies \[ g(y) = \frac {1}{y} f(\ln y), \] as desired.
The mode of \(Y\), \(\lambda \) must satisfy that \(g'(\lambda ) = 0\). By quotient rule, we have \[ g'(y) = \frac {f'(\ln y) \cdot \frac {1}{y} \cdot y - 1 \cdot f (\ln y)}{y^2} = \frac {f'(\ln y) - f(\ln y)}{y^2}. \]
Therefore, \(g'(\lambda ) = 0\) implies that \(f'(\ln \lambda ) = f(\ln \lambda )\) as desired.
This is because it is simply a horizontal shift of \(f(x)\) in the positive \(x\) direction by \(\sigma ^2\) (i.e. this is the integral of \(f(x - \sigma ^2)\)), and this improper integral on \(\RR \) will evaluate to the same value as integrating \(f(x)\), which is simply \(1\).
Expanding the exponent of the integrand gives \begin {align*} - \frac {(x - \mu - \sigma ^2)^2}{2 \sigma ^2} & = - \frac {(x - \mu )^2 + \sigma ^4 - 2 \sigma ^2 (x - \mu )}{2 \sigma ^2} \\ & = -\frac {(x - \mu ^2)}{2 \sigma ^2} - \frac {1}{2}\sigma ^2 + (x - \mu ). \end {align*}
Hence, \begin {align*} \Expt (Y) & = \Expt (e^x) \\ & = \frac {1}{\sigma \sqrt {2\pi }} \int _{-\infty }^{\infty } e^x \cdot e^{-(x - \mu )^2 / (2\sigma ^2)} \Diff x \\ & = \frac {1}{\sigma \sqrt {2\pi }} \int _{-\infty }^{\infty } e^{- (x - \mu )^2 / (2\sigma ^2) + x} \Diff x \\ & = \frac {1}{\sigma \sqrt {2\pi }} \cdot e^{\mu + \frac {1}{2}\sigma ^2} \int _{-\infty }^{\infty } e^{- (x - \mu )^2 / (2\sigma ^2) + x - \frac {1}{2}\sigma ^2 - \mu } \Diff x \\ & = e^{\mu + \frac {1}{2}\sigma ^2} \cdot \frac {1}{\sigma \sqrt {2\pi }} \int _{-\infty }^{\infty } e^{-(x - \mu - \sigma )^2 / (2\sigma ^2)} \Diff x \\ & = e^{\mu + \frac {1}{2}\sigma ^2}, \end {align*}
as desired.
When \(X \sim \Normal (\mu , \sigma ^2)\), \(x_m = \mu \) and therefore \(y_m = e^\mu \). Differentiating the p.d.f. for \(X\) gives \begin {align*} f'(x) & = \frac {1}{\sigma \sqrt {2\pi }} \cdot \frac {-2(x - \mu )}{2\sigma ^2} \cdot e^{-(x - \mu )^2 / (2\sigma ^2)} \\ & = - \frac {x - \mu }{\sigma ^2 \cdot \sigma \sqrt {2\pi }} \cdot e^{-(x - \mu )^2 / (2\sigma ^2)}. \end {align*}
Therefore, \(f(x) = f'(x)\) when \(-\frac {x - \mu }{\sigma ^2} = 1\). This is precisely when \(x = \mu - \sigma ^2\), which means \[ \lambda = e^{\mu - \sigma ^2}. \]
Now, since \(\Expt (Y) = e^{\mu + \frac {1}{2}\sigma ^2}, y_m = e^\mu , \lambda = e^{\mu - \sigma ^2}\), and \(\sigma \neq 0\) so \(\sigma ^2 > 0\), this gives the result \[ \lambda < y_m < \Expt (Y) \] as desired.