Variance and Expected Value Problem

AI Thread Summary
An example of a random variable with an expected value of 4 and variance of 4 can be constructed by first defining a variable A with a mean of 0 and variance of 4, then creating variable B as A+4. This approach ensures that while the mean is adjusted to 4, the variance remains unchanged. The discussion highlights the relationship between expected value and variance, emphasizing that they can be derived from different probability mass functions (p.m.f.). It is noted that E(X) = E(X^2) - (E(X))^2 is not the correct approach for this problem. The focus remains on finding distinct p.m.f.s that yield the same expected value and variance.
dspampi
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Give an example of a random variable (i.e. give the range of values it takes and its p.m.f.) with the following properties: EX = 4, VAR(X)=4. Now give an example of a random variable with a different p.m.f. than the first one you gave, but that still has EX = 4, VAR(X)=4.


So this means then E(X) = E(X^2) - (E(x))^2 right?
I'm not sure if this is the way I should approach the problem?
 
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Not really. They just happen to equal the same number. What I would do is first find a random variable A with a mean of 0 and variance of 4. Then the variable B=A+4 will have a mean of 4 and a variance of 4.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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