0
$\begingroup$

Can somebody help me with the following exercise:

Let $\mathbf{P} \in \mathcal{M}_1\bigl([0,\infty)\bigr)$ with $m_\mathbf{P} := \int x \, \mathbf{P}(dx) \in (0,\infty)$, define a probability measure $\hat{\mathbf{P}}(A) \in \mathcal{M}_1\bigl([0,\infty)\bigr)$ by \begin{equation*} \hat{\mathbf{P}}(A) := \frac{1}{m_\mathbf{P}} \int_A x \, \mathbf{P}(dx), \quad A \in \mathcal{B}\bigl([0,\infty)\bigr). \end{equation*} This is called the size-biased distribution corresponding to $\mathbf{P}$.

Let $(X_i)_{i\in I}$ non-negative real random variables with $\mathbf{E}[X_i] = 1$ for all $i \in I$, $\mathbf{P}_i := \mathbf{P} \circ X_i^{-1}$.

Show that \begin{equation*} \{\hat{\mathbf{P}}_i : i \in I\} \text{ tight } \Longleftrightarrow \{X_i : i \in I\} \text{ uniformly integrable.} \end{equation*}

I need the general idea of the proof or better a full proof (please don't use the "transfer theorem").

Yeah, yeah, "off-topic" & "lack of context" is constantly abused here.

So, please, here's my proof (that pretty much destroys any possibility to get a real answer):

"$\Longrightarrow$":

Assume that $\{\hat{\mathbf{P}}_i : i \in I\}$ are tight. This means that for any $\varepsilon > 0$ there exists a compact set $K\in \mathcal{B}\bigl([0,\infty)\bigr)$, so that \begin{equation*} \sup \bigl\{ \hat{\mathbf{P}}_i\bigl[[0,\infty)\setminus K\bigr] : i \in I \bigr\} < \varepsilon \, . \end{equation*}

Since $K$ is compact, there is an $a \in [0,\infty)$, so that $a > \sup(K)$. Obviously $0 \le \inf(K)$ also holds.

$m_{\mathbf{P}_i} = 1$ because $\mathbf{E}[X_i] = 1$. Also since $X_i$ is non-negative the value of the integral $\int_A X_i \, d\mathbf{P}$ is monotonic regarding $A$. Using these facts we get: \begin{align*} \varepsilon > \hat{\mathbf{P}}_i\bigl[[0,\infty)\setminus K\bigr)\bigr] = \int_{[0,\infty)\setminus K} x \, d\mathbf{P}_i = \int_{X_i^{-1}([0,\infty)\setminus K)} X_i \, d\mathbf{P} \ge \int_{X_i^{-1}((a,\infty))} X_i \, d\mathbf{P} \, . \end{align*}

Since for any $\varepsilon>0$ we can find such an $a \in [0, \infty)$ it must hold that: \begin{equation*} \inf_{a\in[0,\infty)} \sup_{i\in I} \int_{\{|X_i| > a\}} |X_i| \, d\mathbf{P} = 0 \, , \end{equation*} or in other words, the $\{X_i : i\in I\}$ are uniformly integrable.

"$\Longleftarrow$":

Now conversely assume that $\{X_i : i\in I\}$ are uniformly integrable. This means that for any $\varepsilon > 0$ there is an $a\in [0, \infty)$ so that \begin{equation*} \int_{\{|X_i| > a\}} |X_i| \, d\mathbf{P} < \varepsilon \quad \text{ for all }i\in I \, . \end{equation*} The set $K:=[0, a] \subset [0, \infty)$ is compact. Then for any $i\in I$ \begin{align*} \hat{\mathbf{P}}_i\bigl[[0, \infty)\setminus K\bigr] &= \hat{\mathbf{P}}_i\bigl[(a,\infty)\bigr] = \frac{1}{m_{\mathbf{P}_i}} \int_{(a,\infty)} x \, d \mathbf{P}_i \\ &= \frac{1}{m_{\mathbf{P}_i}} \int_{X_i^{-1}((a,\infty))} X_i \, d \mathbf{P} = \int_{\{|X_i| > a\}} |X_i| \, d\mathbf{P} < \varepsilon \, , \end{align*} which means that the $\{\hat{\mathbf{P}}_i : i \in I\}$ are tight. $\square$

$\endgroup$
2
  • $\begingroup$ Just to check, what is $\mathcal{M}?$ $\endgroup$
    – user230715
    May 6 '15 at 8:17
  • $\begingroup$ @GeorgeS: $\mathcal{M}$ = radon measures, $\mathcal{M}_1$ = probability measures $\endgroup$
    – Qyburn
    May 6 '15 at 8:18
1
$\begingroup$

In this context, $m_{\mathbf P_i}=1$ and using the transfer theorem, we have for each $R$, $$\widehat{\mathbf P }_i\left(\mathbf R\setminus [-R,R]\right)=\mathbb E_{\mathbf P}\left[|X_i|\mathbf 1\{|X_i|\gt R\}  \right],$$ where $\mathbf 1(A)$ denotes the indicator function of the set $A$ and $\mathbb E_{\mathbf P}$ the expectation with respect to the probability measure $\mathbf P$.

$\endgroup$
2
  • $\begingroup$ Sorry, I don't understand your notation. For example, what does the $\mathbf{1}$ mean or $\mathbb{E}_{\mathbf{P}}$? $\endgroup$
    – Qyburn
    May 6 '15 at 18:35
  • $\begingroup$ Please see edit. $\endgroup$ May 6 '15 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.