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1

Let $p_{1}, \dots , p_{k} > 1$, be such that $\sum_{j=1}^{k} p_{j}^{-1} = 1$. Let $X_{j}$, $j=1, \dots , k$, be random variables such that $\lVert X_{j} \rVert_{p_{j}} := \Big( E \mid X_{j}\mid^{p_{j}} \Big)^{1/p_{j}} < \infty$. Then show that $E \Big[ \prod_{j=1}^{k} \mid X_{j} \mid \Big] \leq \prod_{j=1}^{k} \lVert X_{j} \rVert_{p_{j}}$.

We will follow the proof of Holder’s inequality.

Notice that by the concavity of $\log$

\[\log \Big( \frac{ a_{1}^{p_{1}} }{ p_{1} } + \dots + \frac{ a_{k}^{p_{k}} }{ p_{k} } \Big) \geq p_{1}^{-1} \log(a_{1}^{p_{1}}) + \dots + p_{k}^{-1} \log(a_{k}^{p_{k}}) = \log(a_{1} + \dots + a_{k}) = \log(a_{1} \cdot \dots \cdot a_{k})\]

Now take $a_{i} = \frac{ \mid X_{i} \mid }{ E[ \mid X_{i} \mid^{p_{i}}]^{1/p_{i}} }$

\[\begin{align} \frac{ \mid X_{1} \mid \cdot \dots \cdot \mid X_{k} \mid }{E[ \mid X_{1} \mid^{p_{1}}]^{1/p_{1}} \cdot \dots \cdot E[ \mid X_{k} \mid^{p_{k}}]^{1/p_{k}} } & \leq \frac{ \mid X_{1} \mid }{ E[ \mid X_{1} \mid^{p_{1}}]^{1/p_{1}} } + \dots + \frac{ \mid X_{k} \mid }{ E[ \mid X_{k} \mid^{p_{k}}]^{1/p_{k}} } \\ \frac{ E [\mid X_{1} \mid \cdot \dots \cdot \mid X_{k} \mid] }{E[ \mid X_{1} \mid^{p_{1}}]^{1/p_{1}} \cdot \dots \cdot E[ \mid X_{k} \mid^{p_{k}}]^{1/p_{k}} } & \leq \frac{ 1 }{ p_{1} } + \dots + \frac{ 1 }{ p_{k} } \\ & = 1 \\ \Rightarrow \\ E[ \mid X_{1} \cdot \dots \cdot X_{k} \mid] & \leq E[ \mid X_{1}^{p_{1}} \mid ]^{1/p_{1}} \cdot \dots \cdot E[ \mid X_{k}^{p_{k}} \mid ]^{1/p_{k}} \end{align}\]

2

Let $X$ and $Y$ be independent random variables, $p \geq 1$, and $E(Y) = 0$. Show that for any $x$, $E \mid Y + x \mid^{p} \geq \mid x \mid^{p}$.

This follows from Jensen’s inequality.

\[E[ \mid Y + x \mid^{p}] \geq \mid E(Y) + x \mid^{P} = \mid x \mid^{p}\]

Show that for $E \mid X \mid^{p} \leq E \mid X+Y \mid^{p}$.

Notice that $E(Y \mid X) = E(Y) = 0$. By Jensen’s inequality, we know that

\[\mid X \mid^{p} = \mid X + E(Y \mid X) \mid^{p} = \mid E(X + Y \mid X) \mid^{p} \leq E( \mid X+Y \mid^{p} \mid X).\]

Then by the law of total expectation,

\[E(\mid X \mid^{p}) \leq E\Big( E( \mid X + Y \mid^{p} \mid X)\Big) = E( \mid X+Y \mid^{p}).\]

3

Show that $\int_{0}^{\infty} e^{-x^{2}/2} dx = \sqrt{ \pi / 2 }$.

Equivalently, we will show that

\[\begin{align} \Big[ \int_{0}^{\infty} & e^{-x^{2}/2} dx \Big]^2 \\ & = \int_{0}^{\infty} e^{-x^{2}/2} dx \cdot \int_{0}^{\infty} e^{-y^{2}/2} dy \\ & = \int_{0}^{\infty}\int_{0}^{\infty} e^{-x^{2}/2} e^{-y^{2}/2} dx dy \\ & = \int_{0}^{\infty}\int_{0}^{\infty} e^{-\frac{ x^{2}+y^{2} }{ 2 }} dx dy \\ & = \frac{ \pi }{ 2 } \end{align}\]

We will perform the polar transformation.

\[\begin{align} \int_{0}^{\infty}\int_{0}^{\infty} & e^{-\frac{ x^{2}+y^{2} }{ 2 }} dx dy\\ & = \int_{0}^{\pi / 2} \int_{0}^{\infty} e^{-R^{2}/2} R dR d\theta \\ & = \frac{ \pi }{ 2 }\int_{0}^{\infty} e^{-R^{2}/2} R dR & \text{Fubini's Theorem} \\ & = \frac{ \pi }{ 2 } \int_{0}^{\infty} e^{-u} R \frac{ du }{ R } & u = \frac{ R^{2} }{ 2 } \\ & = \frac{ \pi }{ 2 } \Big[ -e^{-\infty} - -e^{-0} \Big] \\ & = \frac{ \pi }{ 2 } \end{align}\]

4

If $X$ is a positive random variable and $\alpha > 0$, show that

\[E(X^{\alpha}) = \alpha \int_{[0, \infty)} x^{\alpha-1}P(X > x) dx,\]

where $dx$ in the right hand side stands for integration with respect to the Lebesgue measure.

We can write

\[X = \int_{0}^{X} dx = \int_{0}^{\infty} 1(y < x(\omega)) dy.\]

Then,

\[E(X) = \int_{\Omega} \int_{0}^{\infty} 1(y < X(\omega)) dy d\omega = \int_{0}^{\infty} \int_{\Omega} 1(y < X(\omega)) d\omega dy = \int_{0}^{\infty} P(y < X) dy.\]

Where the change in order of integration comes from Fubini’s Theorem. Also notice we can write

\[X^{\alpha} = \alpha \int_{0}^{X} z^{\alpha-1} dz = \int \alpha x^{\alpha - 1} 1(x < X(\omega)) dx.\]

Then, finally

\[E(X^{\alpha}) = \int_{\Omega} \int_{0}^{\infty} \alpha 1(x^{\alpha-1} < X(\omega)) dx d \omega = \alpha \int_{0}^{\infty} x^{\alpha-1}P(X > x) dx\]

5

For a Borel measurable integration function $f$ on $\mathbb{R}$, show that

\[\int \int f(x-y)f(y) dy dx \geq 0.\]

We can use Fubini’s theorem to swap the order of integration.

\[\int \int f(x-y)f(y) dy dx = \int f(y) \Big[ \int f(x-y) dx \Big] dy\]

Since we are integrating over all the reals, we can re-write the inner integral (notice that the shift will not change the value).

\[\int f(y) \Big[ \int f(x-y) dx \Big] dy = \int f(y) \Big[ \int f(z) dz \Big] dy = \int f(y) dy \cdot \int f(y) dy = \Big[ \int f(y) dy \Big]^{2} \geq 0\]