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):


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.


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. $\square$

  • $\begingroup$ 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

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 $\mathbf P$.

  • $\begingroup$ 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

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