this is for holding javascript data
Anisha Keshavan added I_ve_added_the_full__.tex
over 8 years ago
Commit id: 3f21290b5183f358e2bcb9190305794c869dbbe8
deletions | additions
diff --git a/I_ve_added_the_full__.tex b/I_ve_added_the_full__.tex
new file mode 100644
index 0000000..93c38da
--- /dev/null
+++ b/I_ve_added_the_full__.tex
...
I've added the full derivation to the Appendix, specifically:
The proof for this is found in Introduction to the Theory of Statistics (1974) by Mood, Graybill and Boes \cite{boes1974introduction}, section 2.3, Thoerem 3:
Let $X$ and $Y$ be two random variables where $var[XY]$ exists, then
\begin{equation}
\begin{split}
var[XY] = \mu_Y^2var[X] + \mu_X^2var[Y] + 2\mu_X\mu_Ycov[X,Y]
\\
- (cov[X,Y])^2 + E[(X-\mu_X)^2(Y-\mu_Y)^2]
\\
+ 2\mu_YE[(X-\mu_X)^2(Y-\mu_Y)] + 2\mu_XE[(X-\mu_X)(Y-\mu_Y)^2]
\end{split}
\end{equation}
which can be obtained by computing $E[XY]$ and $E[(XY)^2]$ when $XY$ is expressed as
\begin{equation}
XY = \mu_X\mu_Y + (X-\mu_X)\mu_Y + (Y-\mu_X)\mu_X + (X-\mu_X)(Y-\mu_Y)
\end{equation}
If $X$ and $Y$ are independent, then $E[XY] = \mu_X\mu_Y$, the covariance terms are 0, and
\begin{equation}
E[(X-\mu_X)^2(Y-\mu_Y)^2] = E[(X-\mu_X)^2]E[(Y-\mu_Y)^2] = var[X]var[Y]
\end{equation}
and
\begin{equation}
\mu_YE[(X-\mu_X)^2(Y-\mu_Y)] = E[(X-\mu_X)^2]E[(Y-\mu_Y)] = 0
\end{equation}
\begin{equation}
\mu_XE[(Y-\mu_Y)^2(X-\mu_X)] = E[(Y-\mu_Y)^2]E[(X-\mu_X)] = 0
\end{equation}
Which gives
\begin{equation}
var[XY] = \mu_X^2var[Y] + \mu_Y^2var[X] + var[X]var[Y]
\end{equation}
