Suppose $\displaystyle Y_{n}\ge 0$. Show $\displaystyle Y_{n}\rightarrow 0$ in probability iff $\displaystyle E(\frac{Y_{n}}{1+Y_{n}})\rightarrow 0$.

Any tips would be much appreciated.

Printable View

- Dec 12th 2010, 09:26 PMBeakyconvergence in probability iff expectation...
Suppose $\displaystyle Y_{n}\ge 0$. Show $\displaystyle Y_{n}\rightarrow 0$ in probability iff $\displaystyle E(\frac{Y_{n}}{1+Y_{n}})\rightarrow 0$.

Any tips would be much appreciated. - Dec 13th 2010, 05:23 AMFocus
What have you done so far? Some basic hints, Markov's inequality will be useful and also splitting the expectation.

- Dec 13th 2010, 10:34 AMBeaky
I was trying with Markov's inequality earlier but didn't get anywhere. I think I've managed to show one inclusion now. I don't see how to split the expectation either which is maybe why I'm stuck on the other.

$\displaystyle P(\frac{Y_{n}}{1+Y_{n}}\ge\epsilon)=P(Y_{n}\ge\eps ilon+\epsilon Y_{n})=P(Y_{n}\ge\frac{\epsilon}{1-\epsilon})\le E(\frac{Y_{n}}{1+Y_{n}})/\epsilon$

So I have expectation >> convergence since $\displaystyle \frac{\epsilon}{1-\epsilon}<\epsilon$ for small $\displaystyle \epsilon$. - Dec 13th 2010, 02:43 PMFocus
You are supposed to show that $\displaystyle Y_n \rightarrow 0$ which is pretty similar to what you did if you notice that $\displaystyle \frac{Y_n}{Y_n+1} \leq Y_n$.

For the converse try splitting as follows

$\displaystyle \mathbb{E}[|Y_n/(Y_n+1)|]\leq \mathbb{E}[|Y_n/(Y_n+1)|\mathbf{1}_{|Y_n|>\epsilon}]+\mathbb{E}[|Y_n/(Y_n+1)|\mathbf{1}_{|Y_n|\leq \epsilon}] $

You can bound the RHS term by epsilon (as I indicated above), the second term you can split further by using the fact that $\displaystyle Y_n/(Y_n+1) =1-1/(Y_n+1) \leq 1$ then use convergence in probability.