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Thread: Show Yn->0 in probability iff E[Yn/(1+Yn)]->0

  1. #1
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    Show Yn->0 in probability iff E[Yn/(1+Yn)]->0

    Let Yn≥0.
    a) show that E[Yn/(1+Yn)]->0 => Yn->0 in probability.
    b) show that Yn->0 in probability => E[Yn/(1+Yn)]->0.
    ==========================

    a) Let ε>0.
    Using Markov's inequality,
    P(Yn≥ε) = P{[Yn/(1+Yn)] ≥ ε/(1+Yn)} ≤ E[Yn/(1+Yn)] /[ε/(1+Yn)]
    If we can show that E[Yn/(1+Yn)] /[ε/(1+Yn)]->0, then this implies that Yn->0 in probability and we're done.
    But how can we prove that E[Yn/(1+Yn)] /[ε/(1+Yn)]->0 ? I know that the numerator goes to 0 by assumption, but I am really worrying about the denominator; if Yn->∞ as n->∞, then the denominator would go to 0 as n->∞, and 0/0 is indeterminate!? How can we resolve this problem? I am really scared now, help...


    b) For this part, it seems like the use of Markov's inequality wouldn't work because the inequality goes the "wrong" way, so I am now stuck and I don't know where to start. Could somebody please give me some hints?

    Any help is very much appreciated!
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  2. #2
    Moo
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    Hello,

    Let $\displaystyle Z_n=\frac{Y_n}{1+Y_n}=f(Y_n)$

    Then we want to show that $\displaystyle Y_n\to 0$ in probability iff $\displaystyle E[Z_n]\to 0$

    Note that $\displaystyle f'(y)=\frac{1}{(1+y)^2}>0$

    So f is a strictly increasing bijection from $\displaystyle [0,\infty)$ to $\displaystyle [0,1)$

    So we have $\displaystyle \{Y_n>\epsilon\}=\{f(Y_n)>f(\epsilon)\}=\left\{Z_n >\tfrac{\epsilon}{1+\epsilon}\right\}$ (1)

    And similarly, we have $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ (2)


    a) You indeed have to use Markov's inequality, but I think you did it strangely... because when given P(|X|>a), a should be a constant and not a random variable, like $\displaystyle \frac{\epsilon}{1+Y_n}$ in what you did.

    -----------------------------
    From (1), we have $\displaystyle P(|Y_n|>\epsilon)=P(Y_n>\epsilon)=P\left(Z_n>\tfra c{\epsilon}{1+\epsilon}\right)$, which, by Markov's inequality, is $\displaystyle \leq \frac{1+\epsilon}{\epsilon} \cdot E[Z_n]$

    But we supposed that $\displaystyle E[Z_n]\to 0$. Thus $\displaystyle P(|Y_n|>\epsilon)\to 0~,~ \forall \epsilon>0$

    And this means that $\displaystyle Y_n\to 0$ in probability.



    b) This one is more tricky imo...

    $\displaystyle E[Z_n]=E[Z_n\cdot\bold{1}_{Z_n\leq \epsilon}]+E[Z_n\cdot\bold{1}_{Z_n>\epsilon}]$

    - We can see that $\displaystyle E[Z_n\cdot\bold{1}_{Z_n\leq \epsilon}]\leq E[\epsilon \cdot \bold{1}_{Z_n\leq \epsilon}]=\epsilon \cdot P(Z_n\leq \epsilon)$

    - Note that $\displaystyle Z_n<1$ almost surely (this follows from the property of the function f)
    Hence $\displaystyle E[Z_n\cdot\bold{1}_{Z_n>\epsilon}]<P(Z_n>\epsilon)$
    But by (2), $\displaystyle P(Z_n>\epsilon)=P\left(Y_n>\tfrac{\epsilon}{1-\epsilon}\right)=P(Y_n>\epsilon')$


    So finally, we get $\displaystyle E[Z_n]\leq \epsilon+P(Y_n>\epsilon')$

    Using the fact that $\displaystyle Y_n\to 0$ in probability, $\displaystyle P(Y_n>\epsilon')\to 0$

    And we're left with $\displaystyle 0\leq \lim_{n\to\infty} E[Z_n]\leq \epsilon~,~ \forall \epsilon>0$

    Thus $\displaystyle \lim_{n\to\infty} E[Z_n]=0$
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  3. #3
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    Thank you, Moo! It's actually a lot trickier than I originally thought...

    I have a question about (2): $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$.

    Zn>ε
    <=> Yn/(1+Yn)>ε
    <=> Yn > ε + εYn
    <=> Yn(1-ε) > ε
    <=> Yn > ε/(1-ε) only when ε<1
    What if ε>1??? Then it would be Yn < ε/(1-ε) ???

    So $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ is not true for all ε>0?
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  4. #4
    Moo
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    Quote Originally Posted by kingwinner View Post
    Thank you, Moo! It's actually a lot trickier than I originally thought...

    I have a question about (2): $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$.

    Zn>ε
    <=> Yn/(1+Yn)>ε
    <=> Yn > ε + εYn
    <=> Yn(1-ε) > ε
    <=> Yn > ε/(1-ε) only when ε<1
    What if ε>1??? Then it would be Yn < ε/(1-ε) ???

    So $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ is not true for all ε>0?
    Good remark

    Actually, in general, we need "small" epsilons, as we're interested in what happens when epsilon gets very very small.
    So you can consider $\displaystyle \epsilon <1$ without much loss of generality ^^
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