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Thread: Convergence in probability

  1. #1
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    Convergence in probability

    I guess, it's tough one. I think I should use strong law of large numbers, but I don't know how.
    Random variables $\displaystyle X_{1},X_{2},...$ are independent and $\displaystyle X_{k}$ with gaussian distribution $\displaystyle N(k^{1/2},k)$ k=1,2,..
    Prove that following sequence

    $\displaystyle \frac{1}{n^{2}}(X_{1}X_{2}+X_{3}X_{4}+...+X_{2n-1}X_{2n})$

    is convergent in probability and find the limit.

    Thx for any help.
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  2. #2
    Moo
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    Hello,

    LLN doesn't suit. Yes intuitively the 1/nē is similar to the 1/n in the LLN. But here we're looking for the convergence in probability, which appears in the small law of large numbers, a barely used theorem.
    That's why I think that it isn't the LLN we need here.

    After wandering, I'm thinking of using Chebychev's & Jensen's inequalities. The problem is that nowhere will I use the fact that it's a normal distribution... If you added the normal distribution by some mistake, please tell

    -----------------------------------------

    Let $\displaystyle \displaystyle Z=\frac{1}{n^2}\sum_{k=1}^n X_{2k-1}X_{2k}$

    Chebychev's inequality can be used because $\displaystyle (X_k)$ is an independent sequence with a finite second moment.

    This gives $\displaystyle \displaystyle \forall \epsilon>0,~ P(|Z-E[Z]|\geq \epsilon)\leq \frac{Var[Z]}{\epsilon^2}$


    $\displaystyle \displaystyle n^2 E[Z]=\sum_{k=1}^n E[X_{2k-1}]E[X_{2k}]=\sum_{k=1}^n \sqrt{(2k-1)(2k)}$

    Jensen's inequality is usually for convex functions, but for concave functions you just have to reverse the inequality !

    So since the square root function is concave, we get that $\displaystyle \displaystyle n^2 E[Z]\leq \sqrt{\sum_{k=1}^n 4k^2-2k}$

    But we know that the sum the squares of the first n integers $\displaystyle \sim n^3$ when n gets big. So we can say that $\displaystyle E[Z]\leq a_n \sim \frac{1}{\sqrt{n}}$ as n goes to infinity. If you want to prove it properly well you can do the calculations, but I don't think it's necessary.

    Hence $\displaystyle E[Z]$ tends to 0 as n tends to infinity.

    ~~~~~~~~~~~~~~~~~~~~~~

    Now let's compute $\displaystyle Var[Z]$

    $\displaystyle \displaystyle n^4 Var[Z]=\sum_{k=1}^n Var[X_{2k-1}X_{2k}]$ by independence of the random variables.

    For a given k,
    $\displaystyle \displaystyle \begin{aligned} Var[X_{2k-1}X_{2k}]&=E\left[X_{2k-1}^2 X_{2k}^2\right]-\left[E[X_{2k-1}X_{2k}]\right]^2 \\
    &=(2k-1+2k-1)(2k+2k)-(2k-1)(2k) \\
    &=3(2k-1)(2k) \end{aligned}$

    (I'll let you write the missing steps, I'm not here to do the details )

    So here again we have an equivalent in $\displaystyle n^3$.

    Hence $\displaystyle Var[Z] \sim \frac 1n$ as n goes to infinity.

    ~~~~~~~~~~~~~~~~~~~~~~

    Finally, $\displaystyle \forall \epsilon >0,~ P(|Z-E[Z]|\geq \epsilon) \to 0$ when $\displaystyle n\to\infty$.


    We're almost there. Now we have to prove that Z converges to 0 in probability, by noting that $\displaystyle Z=Z-E[Z]+E[Z]$ (and having in mind that $\displaystyle E[Z]\to 0$ a.s. hence in probability)


    $\displaystyle \displaystyle \forall \epsilon>0,~P(|Z|\geq \epsilon)\leq P(|Z-E[Z]|+E[Z]\geq \epsilon)$.

    - Why ? Because by the triangle inequality, if $\displaystyle |Z|\geq \epsilon$, then $\displaystyle |Z-E[Z]|+E[Z]\geq \epsilon$. Thus $\displaystyle \{\omega~:~ |Z(\omega)|\geq \epsilon\}\subseteq \{\omega~:~|Z(\omega)-E[Z]|+E[Z]\geq \epsilon\}$ and hence the above inequality. -

    Since $\displaystyle |Z-E[Z]|$ and $\displaystyle E[Z]$ are both positive, $\displaystyle \{|Z-E[Z]|+E[Z]\geq \epsilon\}\subseteq \{|Z-E[Z]|\geq \epsilon/2\}\cup \{E[Z]\geq \epsilon/2\}$.

    Hence $\displaystyle \displaystyle \begin{aligned} \forall \epsilon >0,~P(|Z|\geq \epsilon) &\leq P(|Z-E[Z]|+E[Z]\geq \epsilon) \\
    &\leq P(\{|Z-E[Z]|\geq \epsilon/2\}\cup \{E[Z]\geq \epsilon/2\}) \\
    &\leq P(|Z-E[Z]|\geq \epsilon/2)+P(E[Z]\geq \epsilon/2) \\
    &\to 0 \end{aligned}$

    which proves that Z converges to 0 in probability.........
    Last edited by Moo; Jan 17th 2011 at 03:42 AM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    You can probably get almost sure convergence.
    I was looking at

    $\displaystyle \sum_{k=1}^n{Y_k\over k^2}$

    where the Y's are the pair of adjacent normal rvs.
    If you can get that sum to coverge a.s., by the three series theorem
    then by Kronecker you have

    $\displaystyle {\sum_{k=1}^nY_k\over n^2}\to 0$

    But its messy.... showing that

    $\displaystyle \sum_{k=1}^{\infty} P\left( |Y_k|>k^2\right)<\infty$
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  4. #4
    Moo
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    It's impossible to show that for a normal distribution... The result of the problem holds for any distribution (now that I'm sure ) with a finite 2nd moment.
    Then for some distribution, one may get the a.s. convergence, but it would be a problem on its own!
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  5. #5
    MHF Contributor matheagle's Avatar
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    It's impossible to show that

    $\displaystyle \sum_{k=1}^{\infty} P\left( |Y_k|>k^2\right)<\infty$?

    That just involves a double integral, the joint density is known.
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  6. #6
    Moo
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    Have you tried the calculations ?
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  7. #7
    MHF Contributor matheagle's Avatar
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    I don't want to, as I said they are messy.
    One needs to bound the double integral of

    $\displaystyle \int\int_{|xy|<k^2} f(x,y) dA$ for that pair of normals.

    Using the st deviations and means of $\displaystyle \sqrt{2k}$
    But again we don't need to get a precise answer, we only need to obtain bounds
    so that the series converge.
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