We've been given the following problem:

Let $X_1,X_2$, . . . be independent, identically distributed (i.i.d) random variables with $E[X] = 2$ and $var(X)=9$, and let $Y_i = \frac{X_i}{2^i}$. We also define Tn and An to be the sum and the sample mean, respectively, of the random variables Y1, . . . , Yn. In other words, $T_n = \sum_{i = 1}^{n} Y_i$ and $A_n = \sum_{i = 1}^{n} \frac{1}{n}Y_i = \frac{1}{n}T_n$.

The first question is to "Evaluate the mean and variance of Yn, Tn, and An."

I'm a little unsure how to do this without having real numbers. The only thing I can think of is to substitute in the mean and variance of X.

This would give me $E[Y_n] = \frac{2}{2^n}$, $var(Y_n) = \frac{9}{2^n}$ and $E[T_n] = 2(1-1/2^n)$

But this seems like its too simple? and I'm not sure how I'd go about finding the variance in the sums? Would anyone mind possibly pointing me int he right direction?

Thank you!

  • $\begingroup$ I changed $2_i$ to $2^i$ in the definition of $Y_i$. Make sure that is what you intended. $\endgroup$
    – robjohn
    Commented Apr 6, 2016 at 23:32

1 Answer 1


Your calculation of the mean of $Y_n$ is fine.

The variance is not quite right. Recall that the variance of $cW$ is $c^2$ times the variance of $W$.

For calculating the variance of the sums, use the fact that a sum of independent random variables has variance equal to the sum of the individual variances.

  • $\begingroup$ Oh, I didn't know that actually. So that means that $var[Yn] = \frac{1}{2^n} * var[Xi] = \frac{9}{2^n}$? $\endgroup$
    – Indigo
    Commented Apr 6, 2016 at 23:12
  • $\begingroup$ One needs to square the constant. The variance of $Y_n$ is $\left(\frac{1}{2^n}\right)^2\cdot 9$, which can be written as $\frac{9}{2^{2n}}$. $\endgroup$ Commented Apr 6, 2016 at 23:14
  • $\begingroup$ Whoops, I left out the squareing, Thank you $\endgroup$
    – Indigo
    Commented Apr 6, 2016 at 23:16
  • $\begingroup$ So then I just add up all the $var[Yi]$ so $var[Y_n] = \frac{9}{2^{2*1}} +\frac{9}{2^{2*2}} +\frac{9}{2^{2*2}} ... \frac{9}{2^{2n}}$ which you can then sum up using geometric sums, is that right? $\endgroup$
    – Indigo
    Commented Apr 6, 2016 at 23:35
  • $\begingroup$ Note that I had computed the mean of the sum incorrectly, thought that the index started at $0$. The mean is actually $2(1/2+\cdots +1/2^n)$, which simplifies to $2(1-1/2^n)$. And yes, the variance of the sum is what you wrote. One can simplify since it is a geometric series with common ratio $1/4$. $\endgroup$ Commented Apr 6, 2016 at 23:47

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