Sheet 8

Exercise 1 (Complex Integration).

Let $(\mathrm{\Omega },ℱ,\mu )$ be a measure space, $f,g:\mathrm{\Omega }\to ℂ$ an integrable function. Show that

1. ${\int }_{\mathrm{\Omega }}af+g\mathrm{d}\mu =a{\int }_{\mathrm{\Omega }}fd\mu +{\int }_{\mathrm{\Omega }}gd\mu$, for all $a\in ℂ$

   Solution.

   Since $a\in ℂ$, there exist $x,y\in ℝ$ such that $a=x+iy$. Hence we can compute

   ${\displaystyle {\int }_{\mathrm{\Omega }}}af+g\mathrm{d}\mu$ $={\displaystyle {\int }_{\mathrm{\Omega }}}(x+iy)(\mathrm{Re}(f)+i\mathrm{Im}(f))+(\mathrm{Re}(g)+i\mathrm{Im}(g))\mathrm{d}\mu$

   $\stackrel{\text{sort}}{=}{\displaystyle {\int }_{\mathrm{\Omega }}}\underset{\text{real}}{\underbrace{x\mathrm{Re}(f)-y\mathrm{Im}(f)+\mathrm{Re}(g)}}+i\underset{\text{complex}}{\underbrace{(y\mathrm{Re}(f)+x\mathrm{Im}(f)+\mathrm{Im}(g))}}\mathrm{d}\mu$

   $\stackrel{\text{def}}{=}{\displaystyle {\int }_{\mathrm{\Omega }}}x\mathrm{Re}(f)-y\mathrm{Im}(f)+\mathrm{Re}(g)\mathrm{d}\mu +i{\displaystyle {\int }_{\mathrm{\Omega }}}y\mathrm{Re}(f)+x\mathrm{Im}(f)+\mathrm{Im}(g)\mathrm{d}\mu$

   $\stackrel{\text{sort}}{=}(x+iy)\left[{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)d\mu +i{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Im}(f)d\mu \right]+{\displaystyle {\int }_{\mathrm{\Omega }}}\left(\mathrm{Re}(g)+i\mathrm{Im}(g)\right)d\mu$

   $\stackrel{\text{def}}{=}a{\displaystyle {\int }_{\mathrm{\Omega }}}fd\mu +{\displaystyle {\int }_{\mathrm{\Omega }}}gd\mu \mathit{∎}$

2. $\mathrm{Re}({\int }_{\mathrm{\Omega }}fd\mu )={\int }_{\mathrm{\Omega }}\mathrm{Re}(f)d\mu$, $\mathrm{Im}({\int }_{\mathrm{\Omega }}fd\mu )={\int }_{\mathrm{\Omega }}\mathrm{Im}(f)d\mu$

   Solution.

   ${\displaystyle {\int }_{\mathrm{\Omega }}}fd\mu$ $={\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)+i\mathrm{Im}(f)\mathrm{d}\mu$

   $={\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)d\mu +i{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Im}(f)d\mu$

   $=\underset{=\mathrm{Re}\left({\int }_{\mathrm{\Omega }}fd\mu \right)}{\underbrace{{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)d\mu }}+i\underset{=\mathrm{Im}\left({\int }_{\mathrm{\Omega }}fd\mu \right)}{\underbrace{{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Im}(f)d\mu }}\mathit{∎}$

3. $\overline{{\int }_{\mathrm{\Omega }}fd\mu }={\int }_{\mathrm{\Omega }}\overline{f}d\mu$

   Solution.

   $\overline{{\displaystyle {\int }_{\mathrm{\Omega }}}fd\mu }$ $=\overline{\underset{\in ℝ}{\underbrace{{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)d\mu }}+i\underset{\in ℝ}{\underbrace{{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Im}(f)d\mu }}}$

   $={\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Re}(f)d\mu -i{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{Im}(f)d\mu$

   $={\displaystyle {\int }_{\mathrm{\Omega }}}\underset{=\overline{f}}{\underbrace{\mathrm{Re}(f)-i\mathrm{Im}(f)}}d\mu \mathit{∎}$

Exercise 2 (Characteristic Functions).

1. Let $X$ be a random variable on ${ℝ}^d$ with characteristic function ${\phi }_X$. For $a\in ℝ,b\in {ℝ}^d$, show that the characteristic function of $aX+b$ is ${\phi }_{aX+b}(t)={\phi }_X(at)e^{i⟨b,t⟩}$.

   Solution.

   Using (complex) linearity of the integral and bilinearity of the scalar product

   ${\phi }_{aX+b}(t)$ $=𝔼\left[e^{i⟨t,aX+b⟩}\right]\stackrel{\text{bilinear}}{=}𝔼\left[e^{i(a⟨t,X⟩+⟨t,b⟩)}\right]\stackrel{\text{linear}}{=}{e}^{i⟨t,b⟩}𝔼\left[e^{i⟨at,X⟩}\right]$

   $=e^{i⟨t,b⟩}{\phi }_X(at).\mathit{∎}$

2. Let $X$ be a random variable on ${ℝ}^d$ and $Y$ be a random variable on ${ℝ}^k$. Prove that, $X$ is independent of $Y$, if

   $$𝔼\left[\exp \left(i\left(⟨t,X⟩+⟨s,Y⟩\right)\right)\right]={\phi }_X(t){\phi }_Y(s),\forall t\in {ℝ}^d,s\in {ℝ}^k.$$

   Hint.

   The characteristic function determines the distribution, uniquely! Pick a different random vector $\mathrm{(}\stackrel{\mathrm{~}}{X}\mathrm{,}\stackrel{\mathrm{~}}{Y}\mathrm{)}$ with $\stackrel{\mathrm{~}}{X}\stackrel{\mathit{\text{(d)}}}{\mathrm{=}}X$ and $\stackrel{\mathrm{~}}{X}\stackrel{\mathit{\text{(d)}}}{\mathrm{=}}X$ and assume independence.

   Solution.

   Construct a different random vector $(\tilde{X},\tilde{Y})$ with $\tilde{X}\stackrel{\text{(d)}}{=}X$ and similarly $\tilde{Y}$ on a product space to guarantee independence. Then we have due to independence

   $𝔼\left[\exp \left(i\left(⟨t,X⟩+⟨s,Y⟩\right)\right)\right]$ $=𝔼\left[\exp \left(i⟨t,X⟩\exp \right)\cdot \exp \left(i⟨s,Y⟩\right)\right]$

   $\stackrel{⟂⟂}{=}{\phi }_X(t){\phi }_Y(s),\forall t\in {ℝ}^d,s\in {ℝ}^k.$

   Which implies that $(X,Y)$ needs to have the same distribution as $(\tilde{X},\tilde{Y})$. But then they are independent by construction of $\tilde{X},\tilde{Y}$. ∎

Exercise 3 (One-Point Compactification).

We consider the space $([0,\mathrm{\infty }],d)$ with

$d(x,y):=|e^{-x}-e^{-y}|.$

Prove that

1. $d$ is a metric

   Solution.

   We check the requirements:

   • (Positive Definiteness) For $x=y$ we have $d(x,y)=0$ but for $x\ne y$ we have $e^{-x}\ne e^{-y}$ and therefore $d(x,y)>0$.

   • (Symmetry) as $|x|=|-x|$.

   • (Triangle Inequality) By the triangle inequality of the absolute value we have for $x,y,z$

     $d(x,z)$ $=|e^{-x}-e^{-y}+e^{-y}-e^{-z}|\le d(x,y)+d(y,z).\mathit{∎}$

2. and the space is compact.

   Hint.

   Sequences.

   Solution.

   We show that any sequence ${(x_n)}_{n\in \mathbb{N}}\subseteq [0,\mathrm{\infty }]$ has a converging subsequence.

   1. If there is an infinite number of $x_n=\mathrm{\infty }$, then we already have a convergent subsequence. If there is only a finite number, we can only consider the elements afterwards and therefore have w.l.o.g. $(x_n)\subseteq [0,\mathrm{\infty })$.

   2. If the sequence is bounded in $[0,\mathrm{\infty })$ it has a converging subsequence in $[0,\mathrm{\infty })$ which then also converges with regard to $d$ due to continuity of $x\mapsto e^{-x}$. If it is unbounded.

   3. If the sequence is unbounded in $[0,\mathrm{\infty })$, we can find a subsequence such that $x_{n_k}>k$ for all $k$. This implies it converges to $\mathrm{\infty }$ in $[0,\mathrm{\infty }]$ because

      $d(x_{n_k},\mathrm{\infty })$ $=|e^{-x_{n_k}}-\underset{=0}{\underbrace{e^{-\mathrm{\infty }}}}|$

      $\le e^{-n}\to 0.\mathit{∎}$

Exercise 4 (Gaussian Vectors).

For any $k\ge 1$, a random variable $Z:=(Z_1,\mathrm{\dots },Z_k)$ on ${ℝ}^k$ is called a centred Gaussian vector, if every linear combination of $Z_1,\mathrm{\dots },Z_k$ is a centred Gaussian random variable on $ℝ$.

1. Prove that, $Z$ is a centred Gaussian vector, if and only if

   $$𝔼\left[\exp (i⟨t,Z⟩)\right]=\exp \left(-\frac{1}{2}{t}^{T}Mt\right),\forall t\in {ℝ}^k,$$

   where $M$ is a $k\times k$ matrix, with $M_{ij}=\text{Cov}(Z_i,Z_j)$.

   Hint.

   recall that the characteristic function of $𝒩\mathit{}\mathrm{(}\mathrm{0}\mathrm{,}{\sigma }^{\mathrm{2}}\mathrm{)}$ is $\phi \mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}\exp \mathit{}\mathrm{(}\mathrm{-}\frac{\mathrm{1}}{\mathrm{2}}\mathit{}{\sigma }^{\mathrm{2}}\mathit{}{t}^{\mathrm{2}}\mathrm{)}$

   Solution.

   $\Rightarrow$ Suppose that $Z$ is a centered Gaussian random vector. Then $⟨t,Z⟩$ is by definition a centered Gaussian random variable as a linear combination of the $Z_i$. But we know the characteristic function of the univariate Gaussian distribution, i.e. we have

   $𝔼[\exp (i⟨t,Z⟩)]=\exp \left(-{\displaystyle \frac{1}{2}}{\sigma }_t^2\right)$

   where ${\sigma }_t^2$ denotes the variance of $⟨t,Z⟩$. And as it is centered, we have

   ${\sigma }_t^2=𝔼[{⟨t,Z⟩}^2]=𝔼[⟨t,Z⟩⟨Z,t⟩]=𝔼[t^{T}Z{Z}^{T}t]=t^T𝔼[Z{Z}^T]t=t^T\text{Cov}(Z)t=t^{T}Mt.$

   $\Leftarrow$ As the statement holds for all $t$ we can add a parameter and get that

   $𝔼[\exp (i⟨\lambda t,Z⟩)]=\exp \left(-{\displaystyle \frac{1}{2}}{\lambda }^2{t}^{T}Mt\right)\mathit{\hspace{1em}}\forall \lambda \in ℝ.$

   But the first term is the characteristic function in $\lambda$ of $⟨t,Z⟩$. So it is centered Gaussian random variable with variance $t^{T}Mt$ (unique determination of characteristic function). Since this holds for any $t$ (any linear combination) $Z$ is a centered Gaussian vector by definition. ∎

2. Let $(X,Y)$ be a centred Gaussian vector. Prove that, $X$ is independent of $Y$, if and only if $\text{Cov}(X,Y)=0$.

   Solution.

   We only need to show that uncorrelated random variables are independent. By exercise 2(ii) we know that $X$ and $Y$ are independent if and only if

   $𝔼[\exp (i(tX+sY))]={\phi }_X(t)\cdot {\phi }_Y(s)$

   But $\text{Cov}(X,Y)=0$ implies by (i)

   $𝔼[\exp (i(tX+sY))]$

   $=\exp (-\frac{1}{2}[t^2\text{Var}(X)+s^2\text{Var}(Y))$

   $={\phi }_X(t)\cdot {\phi }_Y(s)\mathit{∎}$

Exercise 5 (Complex Analysis).

(Optional Difficult Easter Challenge)

Read about path integrals in Complex Analysis (Funktionentheorie). In particular Cauchy’s Integral Theorem.

1. To calculate an integral over the path ${\gamma }_1:[0,1]\to ℂ,s\mapsto -(n+it)+2sn$ you could therefore find a path ${\gamma }_2$ which connects $n-it$ to $-n-it$. Because connecting both paths together results in a closed loop, we get by Cauchy’s integral theorem

   ${\displaystyle {\int }_{-n}^n}f(y-it)𝑑y\stackrel{x=y-it}{=}{\displaystyle {\int }_{{\gamma }_1}}f(x)𝑑x=-{\displaystyle {\int }_{{\gamma }_2}}f(x)𝑑x.$

   Use this insight to calculate the characteristic function of the standard normal distribution.

   Hint.

   Draw boxes in $ℂ$ considering

   ${\phi }_X(t)$ $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle \int e^{itx}{e}^{-\frac{x^2}{2}}dx}={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle \int e^{-\frac{{(x-it)}^2}{2}}{e}^{-\frac{t^2}{2}}dx}$

   $=e^{-\frac{t^2}{2}}{\displaystyle \frac{1}{\sqrt{2\pi }}}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_{-n}^n}{e}^{-\frac{{(x-it)}^2}{2}}dx.$

   Solution.

   We have

   ${\phi }_X(t)$ $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle \int e^{itx}{e}^{-\frac{x^2}{2}}dx}={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle \int e^{-\frac{{(x-it)}^2}{2}}{e}^{-\frac{t^2}{2}}dx}$

   $=e^{-\frac{t^2}{2}}{\displaystyle \frac{1}{\sqrt{2\pi }}}\underset{n\to \mathrm{\infty }}{lim}{\displaystyle {\int }_{-n}^n}{e}^{-\frac{{(x-it)}^2}{2}}dx,$

   and we therefore get by Cauchy’s Integral Theorem (since $e^{\frac{x}{2}}$ is holomorphic),

   ${\displaystyle {\int }_{-n}^n}{e}^{-\frac{{(x-it)}^2}{2}}dx=-\left(\underset{=:I_1}{\underbrace{{\displaystyle {\int }_t^0}{e}^{-\frac{{(n-is)}^2}{2}}ds}}+\underset{\to -\sqrt{2\pi }(n\to \mathrm{\infty })}{\underbrace{{\displaystyle {\int }_n^{-n}}{e}^{-\frac{x^2}{2}}dx}}+\underset{=:I_2}{\underbrace{{\displaystyle {\int }_0^t}{e}^{-\frac{{(-n-is)}^2}{2}}ds}}\right).$

   So if we can show that ${lim}_n{I}_1=0$ and ${lim}_n{I}_2=0$ we would be finished with

   ${\phi }_X(t)=e^{-\frac{t^2}{2}}.$

   And indeed we have

   $|I_1|$ $\le {\displaystyle {\int }_0^t}\underset{=|e^{-\frac{n^2}{2}}|\underset{=1}{\underbrace{|e^{-ins}|}}|e^{\frac{s^2}{2}}|}{\underbrace{|e^{-\frac{(n^2-2ins-s^2)}{2}}|}}\mathrm{d}s=e^{-\frac{n^2}{2}}\underset{\le t{e}^{t^2/2}}{\underbrace{{\displaystyle {\int }_0^t}{e}^{s^2/2}ds}}\to 0\mathit{\hspace{1em}}(n\to 0).$

   And similarly for $I_2$. ∎

2. Use a similar approach to prove that ${\phi }_X(t)=\frac{1}{1-it}$ for $X\sim \text{Exp}(1)$.

   Hint.

   Here simple boxes will not be enough. A particularly interesting path is the linear path connecting $\mathrm{(}\mathrm{1}\mathrm{-}i\mathit{}t\mathrm{)}\mathit{}a$ to $\mathrm{(}\mathrm{1}\mathrm{-}i\mathit{}t\mathrm{)}\mathit{}b$. Mind the derivative when substituting!

   Solution.

   Call ${\gamma }_{a,b}$ the path linearly connecting $(1-it)a$ to $(1-it)b$. Then we have

   ${\phi }_X(t)$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{itx}{e}^{-x}dx={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-(1-it)x}dx$

   $\stackrel{z=(1-it)x}{=}{\displaystyle \frac{1}{1-it}}{\displaystyle {\int }_{{\gamma }_{0,\mathrm{\infty }}}}{e}^{-z}dz$

   Now by closing the loop we get

   ${\displaystyle {\int }_a^b}{e}^{-y}dy=\underset{=:I_a}{\underbrace{{\displaystyle {\int }_0^{at}}{e}^{-(a-is)}ds}}+{\displaystyle {\int }_{{\gamma }_{a,b}}}{e}^{-y}dy+\underset{=-I_b}{\underbrace{{\displaystyle {\int }_{bt}^0}{e}^{-(b-is)}ds}}.$

   And if we can similarly get rid of the connecting integrals $I_a$ and $I_b$ for $a\to 0$ and $b\to \mathrm{\infty }$, then we immediately get our claim. And this is in fact the case

   $|I_c|$ $=\left|{\displaystyle {\int }_0^{ct}}{e}^{-(c-is)}ds\right|$

   $\stackrel{cx=s}{=}\left|{\displaystyle {\int }_0^t}{e}^{-(c-icx)}cdx\right|$

   $\le c{\displaystyle {\int }_0^t}\underset{=|e^{-c}||e^{-icx}|}{\underbrace{|e^{-(c-icx)|}}}dx$

   $=ct{e}^{-c}\to 0\mathit{\hspace{1em}}\text{for }(c\to 0)\text{ or }(c\to \mathrm{\infty }).\mathit{∎}$