Calculating Variance of Y with X1, X2,...,X15

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The discussion focuses on calculating the variance of the random variable Y, defined as the sum of weighted independent random variables X1 through X15, each taking values +1 or -1 with equal probability. The variance of Y can be derived using the formula VAR(Y) = E(Y^2) - (E(Y))^2. Participants suggest utilizing theorems related to the variance of sums of random variables, emphasizing that the variance of independent variables can be summed directly. The expected value E(Y) is calculated as zero due to symmetry, while E(Y^2) involves summing the squares of the individual components. The final variance of Y is determined to be 175, based on the calculations provided.
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The random variable X1, X2,...,X15 are independent and take each values ​​ +1,-1 with probability 1/2.

We define Y = sum from j=1 to 15(j*Xj)

whats is the variance VAR(Y)=?

i will find this with VAR(Y)=E(X^2)-(E(X)^2) but how i can find them?
 
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ParisSpart said:
The random variable X1, X2,...,X15 are independent and take each values ​​ +1,-1 with probability 1/2.

We define Y = sum from j=1 to 15(j*Xj)

whats is the variance VAR(Y)=?

i will find this with VAR(Y)=E(X^2)-(E(X)^2) but how i can find them?
Your book probably has some theorems that cover finding the variance of a sum of multiple of random variables, such as the following.

If X and Y are random variables, and a and b are constants, then
1. Var(X + Y) = Var(X) + Var(Y) + 2Covar(X, Y)
2. Var(aX) = a2Var(X)
 
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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