Sheet 9

Exercise 1 (Calculating Moments).

Using the characteristic function, calculate the first three moments of

1. $𝒩(\mu ,{\sigma }^2)$

   Solution.

   We have ${\phi }_{𝒩(\mu ,{\sigma }^2)}(t)=\exp (i\mu t-\frac{{\sigma }^2}{2}{t}^2)$ by Example 7.4.12 of the lecture, so by Theorem 7.6.1

   $i𝔼[X]$ $={{\phi }_{𝒩(\mu ,{\sigma }^2)}^{\prime }(t)|}_{t=0}={\exp \left(i\mu t-{\displaystyle \frac{{\sigma }^2}{2}}{t}^2\right)[i\mu -{\sigma }^{2}t]|}_{t=0}=i\mu$

   $i^2𝔼[X^2]$ $={{\phi }_{𝒩(\mu ,{\sigma }^2)}^{\prime \prime }(t)|}_{t=0}={\exp \left(i\mu t-{\displaystyle \frac{{\sigma }^2}{2}}{t}^2\right)({[i\mu -{\sigma }^{2}t]}^2-{\sigma }^2)|}_{t=0}=i^2({\sigma }^2+{\mu }^2)$

   $i^3𝔼[X^3]$ $={{\phi }_{𝒩(\mu ,{\sigma }^2)}^{\prime \prime \prime }(t)|}_{t=0}$

   $={\exp \left(i\mu t-{\displaystyle \frac{{\sigma }^2}{2}}{t}^2\right)({[i\mu -{\sigma }^{2}t]}^3-{\sigma }^2[i\mu -{\sigma }^{2}t]-2{\sigma }^2[i\mu -{\sigma }^{2}t])|}_{t=0}$

   $=i^3({\mu }^3+3{\sigma }^2\mu )\mathit{∎}$

2. $\text{Exp}(\lambda )$

   Solution.

   We have ${\phi }_{\text{Exp}(\lambda )}(t)=\frac{\lambda }{\lambda -it}$ by Example 7.4.12 of the lecture, so by Theorem 7.6.1

   $i𝔼[X]$ $={{\phi }_{\text{Exp}(\lambda )}^{\prime }(t)|}_{t=0}={i\lambda {(\lambda -it)}^{-2}|}_{t=0}=i{\displaystyle \frac{1}{\lambda }}$

   $i^2𝔼[X^2]$ $={{\phi }_{\text{Exp}(\lambda )}^{\prime \prime }(t)|}_{t=0}={2i^2\lambda {(\lambda -it)}^{-3}|}_{t=0}=i^2{\displaystyle \frac{2}{{\lambda }^2}}$

   $i^3𝔼[X^3]$ $={{\phi }_{\text{Exp}(\lambda )}^{\prime \prime \prime }(t)|}_{t=0}={3!i^3\lambda {(\lambda -it)}^{-4}|}_{t=0}=i^3{\displaystyle \frac{3!}{{\lambda }^3}}\mathit{∎}$

3. $\text{Poi}(\lambda )$

   Solution.

   We have ${\phi }_{\text{Poi}(\lambda )}(t)=e^{\lambda (e^{it}-1)}$ by Example 7.4.10 of the lecture, so by Theorem 7.6.1

   $i𝔼[X]$ $={{\phi }_{\text{Poi}(\lambda )}^{\prime }(t)|}_{t=0}={e^{\lambda (e^{it}-1)}\lambda (e^{it})i|}_{t=0}=i\lambda$

   $i^2𝔼[X^2]$ $={{\phi }_{\text{Poi}(\lambda )}^{\prime \prime }(t)|}_{t=0}=e^{\lambda (e^{it}-1)}{(\lambda e^{it}i)}^2+{e^{\lambda (e^{it}-1)}\lambda e^{it}{i}^2|}_{t=0}=i^2({\lambda }^2+\lambda )$

   $i^3𝔼[X^3]$ $={{\phi }_{\text{Poi}(\lambda )}^{\prime \prime \prime }(t)|}_{t=0}={e^{\lambda (e^{it}-1)}[{(\lambda e^{it}i)}^3+2{(\lambda e^{it})}^2{i}^3+{(\lambda e^{it})}^2{i}^3+\lambda e^{it}{i}^3]|}_{t=0}$

   $=i^3({\lambda }^3+3{\lambda }^2+\lambda )\mathit{∎}$

Exercise 2 (Gaussian Vectors 2).

In Sheet 8 Exercise 4 we have defined Gaussian Vectors a little bit different than in the lecture. Here we will show these two definitions are equivalent.

So let $X=\mu +BY$ be a Gaussian vector as defined in the lecture with $Y_i$ independent $𝒩(0,1)$. Since we have talked about centered Gaussian vectors before, let us first deal with the expectation.

1. Prove that $𝔼[X]=\mu$, i.e. $𝔼[X_k]={\mu }_k$

   Solution.

   Let $B_{k\cdot }$ be the $k$-th row of $B$. Then

   $𝔼[X_k]$ $=𝔼[{\mu }_k+⟨B_{k\cdot },Y⟩]={\mu }_k+{\displaystyle \sum _{i=1}^d}{B}_{ki}\underset{=0}{\underbrace{𝔼[Y_i]}}={\mu }_k\mathit{∎}$

2. Prove that for any random vector (not just Gaussian) $X$ the characteristic function of $X+\mu$ is simply

   ${\phi }_{X+\mu }(t)=e^{i⟨t,\mu ⟩}{\phi }_X(t)$

   Solution.

   We can interpret $\mu$ as a random variable and use 7.4.9 (iv) to prove that the characteristic function of the sum is the product of the individual characteristic functions (because a constant is independent from any random variable). So we only need to calculate the characteristic function of a constant

   ${\phi }_{{\delta }_{\mu }}(t)$ $={\displaystyle \int e^{i⟨t,x⟩}d{\delta }_{\mu }(x)}=e^{i⟨t,\mu ⟩}.\mathit{∎}$

Now let us continue on to the (co-)variance and characteristic function.

1. Prove that $\text{Cov}(X_i,X_j)={\mathrm{\Sigma }}_{ij}$ for $\mathrm{\Sigma }:=B{B}^T$.

   Solution.

   We can assume $\mu =0$ without loss of generality since we are subtracting $\mu$ from $X$ in the covariance anyway, so

   $\text{Cov}(X_i,X_j)$ $=𝔼[X_i{X}_j]=𝔼[⟨B_{i\cdot },Y⟩⟨Y,B_{j\cdot }⟩]=B_{i\cdot }\underset{{\mathrm{𝟙}}_{d\times d}}{\underbrace{𝔼[Y{Y}^T]}}{B}_{j\cdot }^T={\mathrm{\Sigma }}_{ij}\mathit{∎}$

2. Prove that $⟨t,X⟩\sim 𝒩(⟨t,\mu ⟩,t^T\mathrm{\Sigma }t)$. (In particular every linear combination is a Gaussian random variable).

   Solution.

   We have

   $⟨t,X⟩=⟨t,\mu ⟩+⟨t,BY⟩=⟨t,\mu ⟩+⟨B^{T}t,Y⟩.$

   The first part is a constant while the second part is a linear combination of independent centered normal random variables. Therefore it is normal with expected value $⟨t,\mu ⟩$. And the variance is

   $𝔼[{⟨B^{T}t,Y⟩}^2]$ $=𝔼[t^{T}BY{Y}^T{B}^{T}t]=t^{T}B{\mathrm{𝟙}}_{d\times d}{B}^{T}t=t^T\mathrm{\Sigma }t.\mathit{∎}$

3. Show that the characteristic function of $X$ is equal to

   ${\phi }_X(t)=\exp (i⟨t,\mu ⟩-\frac{1}{2}⟨t,\mathrm{\Sigma }t⟩).$

   Solution.

   Since we know the distribution of $⟨t,X⟩$ and the characteristic function of normal distributed random variables in one dimension. We have

   ${\phi }_X(t)$ $=𝔼[e^{i⟨t,X⟩}]={\phi }_{⟨t,X⟩}(1)=\exp (i⟨t,\mu ⟩-\frac{1}{2}⟨t,\mathrm{\Sigma }t⟩).\mathit{∎}$

Du to uniqueness of characteristic functions we have shown that the two definitions are equivalent now. Lastly

1. prove that the density of $X$ is given by

   $f_X(x)={\displaystyle \frac{1}{{(2\pi )}^{d/2}det(\mathrm{\Sigma })}}\exp (-{\displaystyle \frac{1}{2}}⟨x-\mu ,{\mathrm{\Sigma }}^{-1}(x-\mu )),\forall x\in {ℝ}^d$

   You may assume $B$ to be invertible.

   Hint.

   Transformation Theorem.

   Solution.

   Let $g$ be any measurable function. Then

   $𝔼[g(X)]$ $=𝔼[g(\mu +BY)]={\displaystyle \frac{1}{{(2\pi )}^{d/2}}}{\displaystyle \int g(\mu +By)\prod _{i=1}^d\exp \left(-\frac{y_i^2}{2}\right)\mathrm{d}y}$

   $={\displaystyle \frac{1}{{(2\pi )}^{d/2}}}{\displaystyle \int g(\mu +By)\exp \left(-\frac{y^{T}y}{2}\right)dy}$

   $={\displaystyle \frac{1}{{(2\pi )}^{d/2}}}{\displaystyle \int g(\mu +By)\exp \left(-\frac{y^T{B}^T{(B^{-1})}^T{B}^{-1}By}{2}\right)dy}$

   $={\displaystyle \frac{1}{{(2\pi )}^{d/2}}}{\displaystyle \int g(\mu +By)\exp \left(-\frac{{(By)}^T{\mathrm{\Sigma }}^{-1}(By)}{2}\right)dy}$

   $={\displaystyle \frac{1}{{(2\pi )}^{d/2}}}{\displaystyle \int g(x)\exp \left(-\frac{{(x-\mu )}^T{\mathrm{\Sigma }}^{-1}(x-\mu )}{2}\right)\frac{1}{det(\mathrm{\Sigma })}dx}\mathit{∎}$

Exercise 3 (Binomial against Poission).

Let $(p_n)$ be a sequence of real numbers in $[0,1]$. Suppose that ${lim}_{n\to \mathrm{\infty }}n{p}_n=\lambda >0$. Using Lévy’s continuity theorem, show that

$$\underset{n\to \mathrm{\infty }}{lim}\left(\genfrac{}{}{0pt}{}{n}{k}\right){(p_n)}^k{(1-p_n)}^{n-k}=e^{-\lambda }\frac{{\lambda }^k}{k!},0\le k\le n.$$

Solution.

From Example 7.4.10 we know that the characteristic functions of Bernoulli and Poisson random variables. So let$X_n\sim \text{Ber}(n,p_n)$ and $X\sim \text{Poi}(\lambda )$, then

${\phi }_{X_n}(t)={\left(1-p_n+p_n{e}^{it}\right)}^n={\left(1+{\displaystyle \frac{n{p}_n{e}^{it}-p_{n}n}{n}}\right)}^n\to \exp (\lambda e^{it}-\lambda )={\phi }_X(t)$

as $n{p}_n\to \lambda$ for $n\to \mathrm{\infty }$. Hence, by Lévy’s continuity theorem 7.5.3 the distributions converge against each other. For

$\underset{n\to \mathrm{\infty }}{lim}\mathbb{P}(X_n=k)\to \mathbb{P}(X=k),$

we want to apply Portmanteau. But (v) is not applicable because the border of the point set $\{X=k\}$ is generally not of probability zero. Instead we view our probability measures ${\mathbb{P}}_{X_N},{\mathbb{P}}_X$ as probability measures on $\mathbb{Z}$. Now using $1/2$-balls it becomes obvious, that all sets on this metric space are open, and therefore all sets are closed. This allows us to use (iii) and (iv) of the Portmanteau theorem for the same result, finishing our proof. ∎

Exercise 4 (Inverse Fourier).

(Optional - Bonus) Assuming measure $\mu$ has density $f=\frac{d\mu }{d\lambda }$, then

$f(x)={\displaystyle \frac{1}{2\pi }}\underset{=:{\widehat{\phi }}_{\mu }(x)}{\underbrace{{\displaystyle \int e^{-itx}{\phi }_{\mu }(t)𝑑t}}}\mathit{\hspace{1em}\hspace{1em}}(\lambda \text{-a.e.})$

for the characteristic function ${\phi }_{\mu }$ of $\mu$.

Hint.

Simply plugging in the definition of the characteristic function and using Fubini won’t work. the result would be this

${\widehat{\phi }}_f(x)={\displaystyle \int e^{-ixt}{\phi }_f(t)dt}={\displaystyle \int e^{-ixt}\int e^{iyt}f(y)dydt}={\displaystyle \int \underset{{\delta }_x(y)}{\underbrace{\int e^{-ixt}{e}^{iyt}dt}}f(y)dy}$

but ${\delta }_x\mathit{}\mathrm{(}y\mathrm{)}$ would need to be a dirac measure on $x$ to have us get back $f\mathit{}\mathrm{(}x\mathrm{)}$ from this point. But Fubini is only allowed for measurable functions and this is not a measurable function. It somehow is zero on all values but $x$, so it is almost surely zero, but at the same time is supposed to integrate to one.

So we have to work around this infinitely high function on one point. And the idea here is to approximate this “delta function” by Gauß kernels

$g_{ϵ}(x)={\displaystyle \frac{1}{\sqrt{2\pi {ϵ}^2}}}\exp (-\frac{x^2}{2{ϵ}^2})\to {\delta }_0(x)\mathit{\hspace{1em}}(ϵ\to 0)$

So you show

${\displaystyle \frac{1}{2\pi }}{\widehat{\phi }}_{\mu }(x)\leftarrow {\displaystyle \frac{1}{2\pi }}{\widehat{\phi }}_{\mu }*g_{ϵ}(x)=f*g_{ϵ}(x)\to f(x)\mathit{\hspace{1em}}ϵ\to 0.$

Here you can plug in definitions and use Fubini freely. Then use the fact that the fourier transform of the normal distribution is basically the density of a normal distribution again to iteratively get rid of all the transforms.

Solution.

We are using a Gauß kernel

$g_{ϵ}(x)={\displaystyle \frac{1}{\sqrt{2\pi {ϵ}^2}}}\exp (-\frac{x^2}{2{ϵ}^2})\to {\delta }_0(x)\mathit{\hspace{1em}}(ϵ\to 0)$

to smudge the equation

${\widehat{\phi }}_{\mu }*g_{ϵ}(x)$ $\stackrel{\text{def.}}{=}{\displaystyle \frac{1}{\sqrt{2\pi {ϵ}^2}}}{\displaystyle \int {\widehat{\phi }}_{\mu }(y)\exp \left(-\frac{{(x-y)}^2}{2{ϵ}^2}\right)dy}$

$\stackrel{\text{def.}}{=}{\displaystyle \frac{1}{\sqrt{2\pi {ϵ}^2}}}{\displaystyle \int \int e^{-ity}{\phi }_{\mu }(t)dt\exp \left(-\frac{{(x-y)}^2}{2{ϵ}^2}\right)dy}$

$={\displaystyle \int \underset{\text{regularization}}{\underbrace{e^{-\frac{1}{2}{t}^2{ϵ}^2}}}\underset{\text{Fourier, vgl. }{\widehat{\phi }}_{\mu }}{\underbrace{e^{-itx}{\phi }_{\mu }(t)}}dt}$

Here we can see that another interpretation of the smushing is that we are using a regularized version of the fourier transform on ${\phi }_{\mu }$. Now if ${\widehat{\phi }}_{\mu }$ is well defined (i.e. if $e^{-itx}{\phi }_{\mu }(t)$ is integrable), then we have $\widehat{\phi }*g_{ϵ}(x)\to \widehat{\phi }(x)$ for $ϵ\to 0$ by the dominated convergence theorem. Continuing on we get

${\widehat{\phi }}_{\mu }*g_{ϵ}(x)$ $\stackrel{\text{def.}}{=}{\displaystyle \int e^{-\frac{1}{2}{t}^2{ϵ}^2}\int e^{itz}f(z)dz}$

$={\displaystyle \int \underset{={\phi }_{𝒩(0,1/{ϵ}^2)}(z-x)\sqrt{2\pi 1/{ϵ}^2}}{\underbrace{\int e^{it(z-x)}{e}^{-\frac{1}{2}{t}^2{ϵ}^2}dt}}f(z)dz}$

$=2\pi {\displaystyle \frac{1}{\sqrt{2\pi {ϵ}^2}}}{\displaystyle \int e^{-\frac{z-x}{2{ϵ}^2}}f(z)dz}$

$=2\pi f*g_{ϵ}(x).$

What is left to do is taking the limit on this side as well. This is a bit tricky. We are going to show that

$\forall f\in L^1:\mathit{\hspace{1em}}f*g_{ϵ}\to f\mathit{\hspace{1em}}\text{in }{L}^1,$ (1)

while the other limit was a pointwise limit! But this is okay due to Fatou’s Lemma

${\displaystyle \int |{\widehat{\phi }}_{\mu }(x)-2\pi f(x)|dx}$ $={\displaystyle \int \underset{ϵ\to 0}{lim\; inf}|{\widehat{\phi }}_{\mu }*g_{ϵ}(x)-2\pi f(x)|\mathrm{d}x}$

$\le 2\pi \underset{=0}{\underbrace{\underset{ϵ\to 0}{lim\; inf}{\displaystyle \int |f*g_{ϵ}(x)-f(x)|dx}}},$

which implies $f=\frac{1}{2\pi }{\widehat{\phi }}_{\mu }$ Lebesgue almost everywhere. But densities are only unique up to Lebesgue zero sets, so we would be done.

Okay so let us show (1): Since $g_{ϵ}$ is a density, we can write

$f*g_{ϵ}(x)-f(x)={\displaystyle \int (f(x-y)-f(x))g_{ϵ}(y)dy}={\displaystyle \int (f(x-ϵz)-f(x))g_1(z)dz},$

where we used the substitution $y=ϵz$ in the last equation. Note that the constant of the normal distribution $\frac{1}{\sqrt{2\pi {ϵ}^2}}$ eats up the $ϵ$ in the substitution $\mathrm{d}y=ϵ\mathrm{d}z$. Therefore we have

${\parallel f*g_{ϵ}-f\parallel }_{L^1(\lambda )}$ $={\displaystyle \int \left|\int (f(x-ϵz)-f(x))g_1(z)dz\right|dx}$

$\stackrel{\text{Fub.}}{=}{\displaystyle \int \underset{=\parallel f(\cdot -ϵz)-f{\parallel }_{L^1(\lambda )}}{\underbrace{\int |f(x-ϵz)-f(x)|dx}}{g}_1(z)dz}.$

Writing ${\tau }_t(f)=f(\cdot -t)$ for the translation by $t$, we have

${\parallel {\tau }_{ϵz}(f)-f\parallel }_{L^1}\le {\parallel {\tau }_{ϵz}(f)\parallel }_{L^1}+{\parallel f\parallel }_{L^1}=2{\parallel f\parallel }_{L^1},$

because a shift does not change the integral over $ℝ$. With this upper bound integrable against $𝒩(0,1)$, we can use the DCT to move the limit into the integral

$\underset{ϵ\to 0}{lim}{\parallel f*g_{ϵ}-f\parallel }_{L^1}={\displaystyle \int \underset{=0}{\underbrace{\underset{ϵ\to 0}{lim}{\parallel {\tau }_{ϵz}(f)-f\parallel }_{L^1}}}{g}_1(z)dz}.$

Where ${lim}_{ϵ\to 0}{\parallel {\tau }_{ϵz}(f)-f\parallel }_{L^1}=0$, because this holds for $f\in C_c$. And, as $C_c$ is dense in $L^1$ (since we can approximate any measurable function with linear combinations of indicators which can in turn be approximated by continuous indicators), it also holds by triangle inequality for all $f\in L^1$. ∎

Exercise 5 (Pólya’s Theorem).

1. Let $X,Y$ be independently $𝒰[-\frac{a}{2},\frac{a}{2}]$ distributed. For $Z=X+Y$ prove that the density is

   $f_Z(x)={\displaystyle \frac{1}{a}}{\left(1-{\displaystyle \frac{|x|}{a}}\right)}^{+}$

   and the characteristic function is

   ${\phi }_Z(t)=2{\displaystyle \frac{1-\cos (at)}{a^2{t}^2}}$

   Hint.

   Lemma 7.4.9. And use the double angle formula

   $\cos (2x)=\cos {(x)}^2-\sin {(x)}^2.$

   Solution.

   The density of a sum of independent random variables is the convolution of their densities:

   $f_Z(x)$ $={\displaystyle \int f_X(x-y)f_Y(y)dy}={\displaystyle \frac{1}{a^2}}{\displaystyle \int {\mathrm{𝟙}}_{[-\frac{a}{2},\frac{a}{2}]}(x-y){\mathrm{𝟙}}_{[-\frac{a}{2},\frac{a}{2}]}(y)𝑑y}$

   $={\displaystyle \frac{1}{a^2}}{\displaystyle \int {\mathrm{𝟙}}_{[-\frac{a}{2},\frac{a}{2}]}(y-x){\mathrm{𝟙}}_{[-\frac{a}{2},\frac{a}{2}]}(y)𝑑y}$

   $={\displaystyle \frac{1}{a^2}}{\displaystyle \int {\mathrm{𝟙}}_{[-\frac{a}{2}+x,\frac{a}{2}+x]}(y){\mathrm{𝟙}}_{[-\frac{a}{2},\frac{a}{2}]}(y)𝑑y}$

   $={\displaystyle \frac{1}{a^2}}{\left(\underset{=\frac{a}{2}-{(-x)}^{+}}{\underbrace{\min \{{\displaystyle \frac{a}{2}}+x,{\displaystyle \frac{a}{2}}\}}}-\underset{=-\frac{a}{2}+{(x)}^{+}}{\underbrace{\max \{-{\displaystyle \frac{a}{2}}+x,-{\displaystyle \frac{a}{2}}\}}}\right)}^{+}$

   $={\displaystyle \frac{1}{a^2}}{\left(a-\underset{=|x|}{\underbrace{[{(-x)}^{+}+{(x)}^{+}]}}\right)}^{+}$

   $={\displaystyle \frac{1}{a}}{\left(1-{\displaystyle \frac{|x|}{a}}\right)}^{+}$

   For the characteristic function we use Lemma 7.4.9 (iv) to get

   ${\phi }_Z(t)$ $={\phi }_X(t)\cdot {\phi }_Y(t)={\left({\displaystyle \frac{\exp (it\frac{a}{2})-\exp (-it\frac{a}{2})}{ita}}\right)}^2$

   $={\left({\displaystyle \frac{\cos (t\frac{a}{2})+i\sin (t\frac{a}{2})-[\cos (-t\frac{a}{2})+i\sin (-t\frac{a}{2})]}{ita}}\right)}^2$

   $={\left({\displaystyle \frac{2\sin (t\frac{a}{2})}{ta}}\right)}^2=4{\displaystyle \frac{\sin {(t\frac{a}{2})}^2}{a^2{t}^2}}=2{\displaystyle \frac{1-\cos (ta)}{a^2{t}^2}}$

   where we have used the trigonometric double angle formula

   $1-\cos (2x)$ $=[\sin {(x)}^2+\underset{=\sin {(x)}^2}{\underbrace{\cos {(x)}^2]-\cos (2x)}}=2\sin {(x)}^2\mathit{∎}$

2. Prove that a random variable $X\sim {\mu }_k$ with density

   $f_{X_k}(x)={\displaystyle \frac{1}{\pi }}{\displaystyle \frac{1-\cos (a_{k}x)}{a_k{x}^2}}$

   has the characteristic function

   ${\phi }_{{\mu }_k}(t)={\left(1-\frac{|t|}{a_k}\right)}^{+}.$

   Deduce the characteristic function ${\phi }_{\mu }$ of $\mu ={\sum }_{k=1}^n{p}_k{\mu }_k$.

   Hint.

   Use (i) and Fourier inversion.

   Solution.

   Using symmetry of the density, we can substitute $x$ with $-x$ to get

   ${\phi }_{{\mu }_k}(t)$ $={\displaystyle \int e^{itx}{f}_{X_k}(x)𝑑x}={\displaystyle \int e^{-itx}{f}_{X_k}(-x)𝑑x}$

   $=a_k{\displaystyle \frac{1}{2\pi }}{\displaystyle \int e^{-itx}\underset{={\phi }_Z(x)}{\underbrace{2\frac{1-\cos (a_{k}x)}{a_k^2{x}^2}}}𝑑x}$

   $=a_k{f}_Z(t)$

   $={\left(1-{\displaystyle \frac{|t|}{a_k}}\right)}^{+},$

   where we have used Fourier inversion in the last inequality with the definition of $Z$ from (i) except using $a_k$ in place of $a$. Since ${\phi }_{{\mu }_k}(0)=1$ this is a probability measure. By linearity of the fourier transform we have

   ${\phi }_{\mu }(t)={\displaystyle \sum _{k=1}^n}{p}_k{\left(1-{\displaystyle \frac{|t|}{a_k}}\right)}^{+}.$

   ∎

3. Assuming $0\le a_1\le \mathrm{\cdots }\le a_n$, calculate the slope of ${\phi }_{\mu }(t)$ on $[a_{k-1},a_k]$. Prove that for given slopes $s_1\le \mathrm{\cdots }\le s_n\le 0=:s_{n+1}$ we can select $\mu$ with $p_k=a_k(s_{k+1}-s_k)$, such that ${\phi }_{\mu }$ has slope $s_k$ on the interval $[a_{k-1},a_k]$ with $a_0=0$.

   Solution.

   For $t\in [a_{k-1}-a_k]$ we can drop the absolute value to get

   ${\displaystyle \frac{d}{dt}}{\phi }_{\mu }={\displaystyle \frac{d}{dt}}{\displaystyle \sum _{i=1}^n}{p}_i{\left(1-{\displaystyle \frac{t}{a_i}}\right)}^{+}={\displaystyle \frac{d}{dt}}{\displaystyle \sum _{i=k}^n}{p}_i\left(1-{\displaystyle \frac{t}{a_i}}\right)=-{\displaystyle \sum _{i=k}^n}{\displaystyle \frac{p_i}{a_i}}.$

   So for $p_k=a_k(s_{k+1}-s_k)$ we have

   ${\displaystyle \frac{d}{dt}}{\phi }_{\mu }$ $=-{\displaystyle \sum _{i=k}^n}{\displaystyle \frac{a_i(s_{i+1}-s_i)}{a_i}}=-(s_{n+1}-s_k)=s_k.\mathit{∎}$

4. Prove Pólya’s Theorem using interpolation points $a_k$, previous results and Lévy’s Continuity Theorem.

   Theorem (Pólya).

   Let $f\mathrm{:}ℝ\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$ be a function which is continuous, even, convex on $\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}$ and satisfies $f\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{1}$ and ${\lim }_{x\mathrm{\to }\mathrm{\infty }}\mathit{}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}\mathrm{0}$. Then $f$ is a characteristic function for some probability measure.

   Solution.

   As $f$ is convex on $[0,\mathrm{\infty })$ the negative slopes between the interpolation points are increasing (decreasing in absolute value). Therefore the $p_k=a_k(s_{k+1}-s_k)$ are positive. So the resulting $\mu$ is a measure. If we can show

   $f(a_k)={\phi }_{\mu }(a_k)\mathit{\hspace{1em}}\forall k=0,\mathrm{\dots },n$

   we get the fact, that $\mu$ is a probability measure for free, as then ${\phi }_{\mu }(0)=f(0)=1$. And by increasing the number of interpolation points we get convergence against $f$, so $f$ would be a characteristic function by Lévy’s continuity theorem. Okay, so let us try to prove this equality

   ${\phi }_{\mu }(a_l)$ $={\displaystyle \sum _{k=l}^n}\stackrel{=p_k}{\overbrace{a_k(s_{k+1}-s_k)}}{\left(1-{\displaystyle \frac{a_l}{a_k}}\right)}^{+}={\displaystyle \sum _{k=l}^n}(s_{k+1}-s_k){(a_k-a_l)}^{+}$

   $=[(a_n-a_l)\underset{=0}{\underbrace{s_{n+1}}}-(a_l-a_l)s_l]-{\displaystyle \sum _{k=l+1}^n}{s}_k(a_k-a_{k-1})$

   $\stackrel{\text{def. slope}}{=}-{\displaystyle \sum _{k=l+1}^n}f(a_k)-f(a_{k-1})$

   $=f(a_l)\mathit{∎}$