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kjetil b halvorsen
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Hi. I'm trying to solve the second statement of the following exercise. It is Exercise 2.5 of "All of Nonparametric Statistics, Larry Wasserman".

enter image description here

My try:

$|T(F)-T(G)| = |\int x dF - \int x dG| = |\int x d(F-G)|$ $\leq \int |x|d(|F-G|) \leq M \int d(|F-G|) \leq M \sup_{x}|F(x)-G(x)|$\begin{align} |T(F)-T(G)| &= \left|\int x dF - \int x dG\right| \\ &= \left|\int x d(F-G)\right| \\ &\leq \int |x|d(|F-G|) \\ &\leq M \int d(|F-G|) \\ &\leq M \sup_{x}|F(x)-G(x)| \end{align}

However, I can not verify the last inequality, rather is. Is it true?

This is Exercise 2.5 of "All of Nonparametric Statistics, Larry Wasserman"

enter image description here

Hi. I'm trying to solve the second statement.

My try:

$|T(F)-T(G)| = |\int x dF - \int x dG| = |\int x d(F-G)|$ $\leq \int |x|d(|F-G|) \leq M \int d(|F-G|) \leq M \sup_{x}|F(x)-G(x)|$

However, I can not verify the last inequality, rather is it true?

This is Exercise 2.5 of "All of Nonparametric Statistics, Larry Wasserman"

I'm trying to solve the second statement of the following exercise. It is Exercise 2.5 of "All of Nonparametric Statistics, Larry Wasserman".

enter image description here

My try:

\begin{align} |T(F)-T(G)| &= \left|\int x dF - \int x dG\right| \\ &= \left|\int x d(F-G)\right| \\ &\leq \int |x|d(|F-G|) \\ &\leq M \int d(|F-G|) \\ &\leq M \sup_{x}|F(x)-G(x)| \end{align}

I can not verify the last inequality. Is it true?

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expectation functional property

enter image description here

Hi. I'm trying to solve the second statement.

My try:

$|T(F)-T(G)| = |\int x dF - \int x dG| = |\int x d(F-G)|$ $\leq \int |x|d(|F-G|) \leq M \int d(|F-G|) \leq M \sup_{x}|F(x)-G(x)|$

However, I can not verify the last inequality, rather is it true?

This is Exercise 2.5 of "All of Nonparametric Statistics, Larry Wasserman"