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Math Help - 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 Z_n=\frac{Y_n}{1+Y_n}=f(Y_n)

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

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

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

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

    And similarly, we have \{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 \frac{\epsilon}{1+Y_n} in what you did.

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

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

    And this means that Y_n\to 0 in probability.



    b) This one is more tricky imo...

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


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

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

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

    Thus \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): \{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 \{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): \{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 \{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 \epsilon <1 without much loss of generality ^^
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