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Thread: Approximating Uniform Dist by CLT

  1. #1
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    Approximating Uniform Dist by CLT

    I'm given a question which asks the following:

    In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, $\displaystyle X_{i}$, $\displaystyle i = 1,...,n,$ are independently distributed as $\displaystyle U(-0.5,+0.5)$

    Obtain the approximate distribution of the total error $\displaystyle \sum X_{i}$ in the sum of the n numbers and hence find the probability that the absolute error in the sum is at most $\displaystyle \frac{1}{2}\sqrt{n}$.

    Now, I assume it wants me to approx to Normal dist by CLT, which is:

    $\displaystyle
    \frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}$ ~ $\displaystyle N(0,1)$

    From the Uniform dist, I know E(X) = 0, and Var(X) = $\displaystyle \frac{1}{12}$

    So does this mean that the approximate dist of $\displaystyle \sum X_{i}$ is

    $\displaystyle \overline{X_{i}}\sqrt{12n}$ ~ $\displaystyle N(0,1)$

    or am I missing something regarding the fact that it's the dist of the sum rather than just X itself?
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  2. #2
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    Quote Originally Posted by Zenter View Post
    I'm given a question which asks the following:

    In adding n real numbers, each is rounded to the nearest integer. Assume that the round-off errors, $\displaystyle X_{i}$, $\displaystyle i = 1,...,n,$ are independently distributed as $\displaystyle U(-0.5,+0.5)$

    Obtain the approximate distribution of the total error $\displaystyle \sum X_{i}$ in the sum of the n numbers and hence find the probability that the absolute error in the sum is at most $\displaystyle \frac{1}{2}\sqrt{n}$.

    Now, I assume it wants me to approx to Normal dist by CLT, which is:

    $\displaystyle
    \frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}$ ~ $\displaystyle N(0,1)$

    From the Uniform dist, I know E(X) = 0, and Var(X) = $\displaystyle \frac{1}{12}$

    So does this mean that the approximate dist of $\displaystyle \sum X_{i}$ is

    $\displaystyle \overline{X_{i}}\sqrt{12n}$ ~ $\displaystyle N(0,1)$

    or am I missing something regarding the fact that it's the dist of the sum rather than just X itself?
    Total error $\displaystyle X=\sum X_i \sim N(0,n/12)$ (approx, by CLT)

    (Note: $\displaystyle \overline{X_{i}}=0$)

    CB
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  3. #3
    MHF Contributor matheagle's Avatar
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    $\displaystyle \mu=0$

    CB uses $\displaystyle \overline{X}$ as the population mean

    The sample mean $\displaystyle \overline{X}$ is not zero, it's a random variable and there is no subscript i to it.
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  4. #4
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    Quote Originally Posted by matheagle View Post
    $\displaystyle \mu=0$

    CB uses $\displaystyle \overline{X}$ as the population mean

    The sample mean $\displaystyle \overline{X}$ is not zero, it's a random variable and there is no subscript i to it.
    There appears to be some part of the conversation missing here, my comment about $\displaystyle \overline X_i$ is due to its appearence in the last expression in the original post, so the LHS of that expression is a numeric constant equal to zero and not a standard normal RV.

    $\displaystyle X_i$ is the RV representing the $\displaystyle i$-th round off error and is $\displaystyle \sim U(-0.5,0.5)$ so by definition $\displaystyle \overline X_i =E( X_i)=0$

    CB
    Last edited by CaptainBlack; Aug 25th 2009 at 11:04 PM.
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Total error $\displaystyle X=\sum X_i \sim N(0,n/12)$ (approx, by CLT)

    (Note: $\displaystyle \overline{X_{i}}=0$)

    CB
    I understand what you did here, but then using this for the part where I work out the probability, I get

    $\displaystyle P(Z \leq \frac{\sqrt{12n}}{2})$

    Which, I don't think would be the right answer, since I'm not given a numerical figure for n, meaning I can't work that ^ out.
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  6. #6
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    Quote Originally Posted by Zenter View Post
    I understand what you did here, but then using this for the part where I work out the probability, I get

    $\displaystyle P(Z \leq \frac{\sqrt{12n}}{2})$

    Which, I don't think would be the right answer, since I'm not given a numerical figure for n, meaning I can't work that ^ out.
    That's funny, 'cos I get $\displaystyle \mathrm P(|Z|<\surd3)$. Any other offers?
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  7. #7
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    Quote Originally Posted by halbard View Post
    That's funny, 'cos I get $\displaystyle \mathrm P(|Z|<\surd3)$. Any other offers?
    Which is what I get as well, the $\displaystyle \sqrt{n}$ terms all cancel.

    CB
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