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Math Help - convergence in probability iff expectation...

  1. #1
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    convergence in probability iff expectation...

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

    Any tips would be much appreciated.
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  2. #2
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    What have you done so far? Some basic hints, Markov's inequality will be useful and also splitting the expectation.
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  3. #3
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    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.

    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 \frac{\epsilon}{1-\epsilon}<\epsilon for small \epsilon.
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  4. #4
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    Quote Originally Posted by Beaky View Post
    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.

    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 \frac{\epsilon}{1-\epsilon}<\epsilon for small \epsilon.
    You are supposed to show that Y_n \rightarrow 0 which is pretty similar to what you did if you notice that  \frac{Y_n}{Y_n+1} \leq Y_n.

    For the converse try splitting as follows
    \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 Y_n/(Y_n+1) =1-1/(Y_n+1) \leq 1 then use convergence in probability.
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