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Let $X, Y$ be independent random variables such that $X,Y \sim N(\mu,\sigma^2)$, show that $X+Y$ and $X-Y$ are independent using the moment generating function.

I know that the moment generating function of a sum of independent random variables is the product of the MGF.

So, I´m trying to solve that but i don´t know if my process is correct

$M_{X+Y}(t_1,t_2)=M_X(t_1)M_Y(t_2)=M^2_{N(\mu,\sigma^2)}(t)$ ?

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    $\begingroup$ What you need to show is the MGF for the random vector $(X + Y, X - Y)$ can be factorized to the product of MGFs of $X + Y$ and $X - Y$. $\endgroup$ – Zhanxiong Jan 25 '16 at 0:44
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Recall that for an $\mathcal{N}(\mu, \sigma^2)$ random variable, the moment generating function of it is $$M(t) = \exp\left(\mu t + \frac{1}{2}\sigma^2t^2\right). \tag{1}$$

By condition, $X + Y \sim \mathcal{N}(2\mu, 2\sigma^2)$ and $X - Y \sim \mathcal{N}(0, 2\sigma^2)$. Therefore by $(1)$, we have: $$M_{X + Y}(t) = \exp\left(2\mu t + \sigma^2 t^2\right), \; M_{X - Y}(t) = \exp\left(\sigma^2 t^2\right).$$

On the other hand, as a bivariate random vector $(X + Y, X - Y)$, its MGF can be computed by definition as follows: \begin{align} & M_{(X + Y, X - Y)}(t_1, t_2) \\ = & E[\exp(t_1(X + Y) + t_2(X - Y))] \\ = & E\left\{\exp[(t_1 + t_2)X] \times \exp[(t_1 - t_2)Y]\right\} \\ = & E\left\{\exp[(t_1 + t_2)X] \right\}\times E\left\{\exp[(t_1 - t_2)Y]\right\} \quad \text{by independence of $X$ and $Y$.}\\ = & M_X(t_1 + t_2) M_Y(t_1 - t_2) \\ = & \exp\left(\mu(t_1 + t_2) + \frac{1}{2}\sigma^2(t_1 + t_2)^2\right)\exp\left(\mu(t_1 - t_2) + \frac{1}{2}\sigma^2(t_1 - t_2)^2\right) \\ = & \exp\left(2\mu t_1 + \sigma^2 t_1^2\right)\exp\left(\sigma^2 t_2^2\right) \\ = & M_{X + Y}(t_1) M_{X - Y}(t_2). \end{align}

Hence $X + Y$ and $X - Y$ are independent.

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  • $\begingroup$ I have one question, when you say: $E\left\{\exp[(t_1 + t_2)X] \right\}\times E\left\{\exp[(t_1 - t_2)Y]\right\} \quad \text{by independence of $X$ and $Y$.}\\$ it is a property?, I only know that E(XY)=E(X)E(Y) but I didn´t know that property. $\endgroup$ – A P Jan 25 '16 at 1:24
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    $\begingroup$ Yes, it is exactly this property, where you treat $(t_1 + t_2)X$ and $(t_1 - t_2)Y$ are independent random variables. Notice that $t_1$ and $t_2$ are all constants. $\endgroup$ – Zhanxiong Jan 25 '16 at 1:36
  • $\begingroup$ so, when I have 2 independent random variables X,Y $E[g(aX)h(bY)]=E[g(aX)]\cdot E[h(bY)]$ is always true? $\endgroup$ – A P Jan 25 '16 at 1:44
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    $\begingroup$ Yes, you made a good guess. Of course, rigorous, $g$ and $h$ needs to be measurable, which is usually guaranteed. $\endgroup$ – Zhanxiong Jan 25 '16 at 1:45
  • $\begingroup$ it´s an interesant result that i didn´t know, can you suggest me a site where find the proof to this property? $\endgroup$ – A P Jan 25 '16 at 1:52
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No. If $X$ and $2 Y = X$, if $X$ is normally distributed so is $Y$. And $X + Y = 3 Y$ and $X - Y = Y$ aren't independent at all.

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  • $\begingroup$ OP did mentioned independence of $X$ and $Y$ $\endgroup$ – Francis Jan 25 '16 at 0:44
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An answer using moment generating functions has already been given, but please also consider the following simpler approach. Note that two jointly normal random variables are independent if and only if they are uncorrelated. Since $X,Y$ are independent normals, the pair $(X,Y)$ is normal. Any linear transformation of a normal is also a normal, so $(X+Y, X-Y)$ is normal, i.e. $X+Y$ and $X-Y$ are jointly normal. Then $E[(X+Y)(X-Y)] = E[X^2 - Y^2] = \sigma^2 - \sigma^2 = 0 = (2 \mu) \cdot 0 = E[X+Y]E[X-Y]$. Thus $X+Y$ and $X-Y$ are independent.

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