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Thread: Uniform Distribution Problem

  1. #1
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    Uniform Distribution Problem

    I am confused as to when finding the mean, why do we use the transformation: Y= (N+2)/2..... since the formula for uniform distribution is (N+1)/2

    Moreover, how does the var[X] = 4 * var [y]
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  2. #2
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    Re: Uniform Distribution Problem

    The formula for the mean of the uniform distribution with 1, 2, 3, …, n, all equally likely, is (n+1)/2. The uniform distribution given here s 0, 2, 4, …, 22. The transformation Y= (X+ 2)/2 converts that to the previous distribution: when X= 0, Y= (0+ 2)/2= 1, when X= 2, Y= (2+ 2)/2= 2, etc. You can then apply the formula $\displaystyle \mu= (n+ 1)/2$ with n= 11.

    As for the variance, that is defined as the average of $\displaystyle (X- \mu)^2$. Since X is two times Y (the "+2" cancels), The variance of X is $\displaystyle 2^2= 4$ times the variance of Y.
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  3. #3
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    Re: Uniform Distribution Problem

    Is there anyway you can expound on the variance of x part? I know that E[X^2] = 506/3 and E[X]=11 which makes (E[X])^2=121. Which E[x^2]-(E[X])^2 = 143/3, but I am not understanding how we find E[X^2].

    I also tried plugging in E[X-(2y-2)]^2 or 11 and I cannot get the same answer.
    Last edited by math951; Oct 19th 2018 at 05:14 PM.
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  4. #4
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    Re: Uniform Distribution Problem

    The data given are 0, 2, 4, , 22, each with probability 1/12. The direct way to calculate the mean is to add all of those and divide by 12: (0+ 2+ 4+ + 22)/12= 132/12= 11. The "Y" transformation just changed to a distribution for which you already knew the mean. To find the variance, the expected value of (x- mean)^2, first subtract 11 from each value: -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11. Now square: 121, 81, 49, 25, 9, 1, 1, 9, 25, 49, 81, 121. To find the variance of X, find the mean of those- add them (you can just add 121+ 81+ 49+ 25+ 9+ 1 and double it) and divide by 12: 572/12= 143/3.

    You are using a simplified formula for the variance. The variance is defined, and I used the definition above, as $\displaystyle E(x- \mu)^2$. For a uniform distribution like this one, we have $\displaystyle \frac{1}{n}\sum_{i=1}^n (X- \mu)^2= \frac{1}{n}\sum_{i=1}^n(X^2- 2\mu X+ \mu^2)$. We can break that up into three sums: $\displaystyle \frac{1}{n}\sum_{i=1}^n X^2- 2\mu \frac{1}{n}\sum_{i=1}^n X+ \mu^2\frac{1}{n}\sum_{i=1}^n 1$. But $\displaystyle \frac{1}{n}\sum_{i=1}^n X$ is the mean, $\displaystyle \mu$ and $\displaystyle \sum_{i=1}^n 1$= n[/tex] so that $\displaystyle \frac{1}{n}\sum_{i=1}^n 1= 1$. So $\displaystyle E((X- \mu)^2)= E(X^2)- 2\mu^2+ \mu^2= E(x^2)- \mu^2$.

    And $\displaystyle E(X^2)$ is just the expected value of $\displaystyle X^2$! With the original data, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, the squares are 0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484. Adding those, $\displaystyle E(X^2)= \frac{2024}{12}= \frac{506}{3}$. Then $\displaystyle E(X^2)- \mu^2= \frac{506}{3}- 121= \frac{506- 363}{3}= \frac{143}{3}$ as before.
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