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Sum of two independent Normal random variables 11/25

Suppose we consider $U \sim N(0,1)$ and $V \sim N(0,1)$, both indepedent, I showed that through rotational symmetry of the joint distribution of (U,V) any change of coordinate system, by a rotation of the first axis for instance would also give a Normal(0,1) random variable:

\begin{displaymath}X_\theta=U cos \theta +V sin \theta\end{displaymath}

thus for any numbers $\alpha$ and $\beta$ such that $\alpha^2+\beta^2=1$, we take the angle $\theta$ such that $\alpha=cos \theta$ and $\beta=sin \theta$, and the linear combination of $W=\alpha U +\beta V= U cos \theta +V sin \theta$ will be Normal(0,1).

As a consequence, the sum of two independent standard Normals will be Normal(0,2). (Taking $\alpha=\beta=\frac{1}{\sqrt{2}})$

More generally suppose we want to consider the sum of two independent Normals that are not standardized:

\begin{displaymath}X\sim N(\mu, \sigma^2) \qquad Y \sim N(\nu, \tau^2)\end{displaymath}

We know through the chapter on expectations and variances that the sum of these two independent random variables will have expectation $ (\mu+\nu)$ and variance $\sigma^2+\tau^2$. Now we consider the standardized random variable

\begin{displaymath}W=\frac{X+Y-
(\mu+\nu)}
{\sqrt{\sigma^2+\tau^2}}\end{displaymath}

this can be rewritten as the sum :

\begin{displaymath}W=\frac{X-\mu}{\sigma}\times \frac{\sigma}
{\sqrt{\sigma^2+\t...
...} +\frac{Y-\nu}{\tau}\times\frac{\tau}{\sqrt{\sigma^2+\tau^2}}
\end{displaymath}

which is of the form $\alpha U +\beta V$, with $\alpha^2+\beta^2=1$, and U and V independent standard normals.
next up previous index
Next: Limit Theorems Up: Sums of Continuous Random Previous: Gamma density
Susan Holmes
1998-12-07