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I studying for a final and I am going questions in old exam for practice. I am having trouble with a question.

"Suppose that $X~$Uniform$[0,12]$ and $Y~$Uniform$[0,12]$ are independent random variables. Define $S=X+Y$ to be the sum of the two independent uniform random variables. Determine the variance of $S$?"

I know that I am supposed to use the Convolution formula 

      integrate fx(x)*fy(s-x)dx from 0 to s

However I don't know how to "setup" the problem. Specifically the $fy(s-x)$ part. I know I can't say $E[S]-E[X]$ because the formula has lowercase $s$ and lowercase $x$). Can some please show me how to set up the problem using the formula. I would like to answer this question using the convolution formula because I going to be expected to know how to apply and understand it for my final. I know that since X and Y are over the same interval it is probably not necessary but I would still like to know how to use it.

rex1637
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  • Well, to start with, your convolution formula is wrong. It should read $f_S(s)=\int_{-\infty}^\infty f_X(x) f_Y(s-x),dx.$ –  Dec 13 '15 at 17:13
  • According to my textbook, for the case S=X+Y, the limits of integration are from 0 to s. I'm looking at it right now and that what the formula is showing. – rex1637 Dec 13 '15 at 17:44
  • That works when the random variables $X$ and $Y$ are non-negative, but not always. On the other hand, I'll bet the formula doesn't integrate the function $f_S(x)$, but rather the function $f_X(x).$ –  Dec 13 '15 at 17:48
  • yes you are right I have edited the question to reflect this. – rex1637 Dec 13 '15 at 17:54
  • Andre's answer here explains convolution very well: http://math.stackexchange.com/questions/357672/density-of-sum-of-two-uniform-random-variables-0-1 –  Dec 14 '15 at 01:31

1 Answers1

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It is likely that you are expected to know and use the fact that the variance of a sum of two independent random variables is the sum of the variances.

So we only need to compute the variance of say $X$. The mean of $X$ is $6$ by symmetry. Now use the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. We need to compute $E(X^2)$, which is $\int_0^{12}\frac{x^2}{12}\,dx$.

André Nicolas
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