# If $X_1$ and $X_2$ are independent $\mathcal N(0,1)$ random variables, then $\frac{X_1-X_2}{\sqrt2}\sim \mathcal N(0,1)$

I am trying to show that if $X_1$ and $X_2$ are independent $\mathcal N(0,1)$ random variables, then $$Z=\dfrac{X_1-X_2}{\sqrt{2}}\sim \mathcal N(0,1)$$

My attempt: Find the moment generating function: \begin{align}M_Z(t)&=E\left[\exp\left(t\dfrac{X_1-X_2}{\sqrt2}\right)\right]=E\left[\exp\left(\dfrac{tX_1}{\sqrt2}\right)\right]E\left[\exp\left(\dfrac{-tX_2}{\sqrt 2}\right)\right]\\[0.2cm]&=-2E\left[\exp(tX_1)\right]E\left[\exp({tX_2})\right]=-2M_{X_1}(t)M_{X_2}(t)=-2\exp\left(\dfrac{1}{2}t^2\right)\end{align}

Is this correct so far? How can I proceed?

• The step $$E[exp(\dfrac{tx_1}{\sqrt 2})]E[exp(\dfrac{-tx_2}{\sqrt 2})]=-2E[exp(tx_1)]E[exp({tx_2})]$$ is (a real mystery and) incorrect. – Did Nov 4 '16 at 6:33

Normal distribution is uniquely determined by its moments, hence, indeed, it suffices to show that the moment generating function (MGF) of $Z$ coincides with the MGF of $\mathcal N(0,1)$ which is equal to $\displaystyle\exp\left(\frac12t^2\right)$. Now, using the following property

If $\displaystyle S_{n}=\sum _{i=1}^{n}a_{i}X_{i}$, where the $X_i$ are independent random variables and the $a_i$ are constants, then the moment-generating function for $S_n$ is given by $$M_{S_{n}}(t)=M_{X_{1}}(a_{1}t)M_{X_{2}}(a_{2}t)\cdots M_{X_{n}}(a_{n}t)$$

you obtain for $Z=\frac1{\sqrt2}X_1+\left(\frac{-1}{\sqrt2}\right)X_2$ that \begin{align}M_Z(t)&=M_{X_1}\left(\frac{t}{\sqrt2}\right)M_{X_2}\left(\frac{-t}{\sqrt2}\right)=\exp\left(\frac12\cdot\left(\frac{t}{\sqrt2}\right)^2\right)\exp\left(\frac12\cdot\left(\frac{-t}{\sqrt2}\right)^2\right)\\[0.2cm]&=\exp\left(\frac{t^2}4\right)\exp\left(\frac{t^2}4\right)=\exp\left(\frac12{t^2}\right)\end{align} which concludes the proof.

For a normal random variable X obviously -X is also normal. The sum of two independent normal random variables ist also normal.

So you already know that $\frac{X_1 - X_2}{\sqrt{2}}$ is normal, with expectation

$$E[\frac{X_1 - X_2}{\sqrt{2}}] = \frac{E[X_1] - E[X_2]}{\sqrt{2}} = 0$$

and variance

$$Var[\frac{X_1 - X_2}{\sqrt{2}}] = \frac{Var[X_1] + Var[X_2]}{2} = 1$$

So it's $\mathcal N(0,1)$

• can the variance be distributed?@Gono – MAS Nov 4 '16 at 10:35
• For independent rv it holds Var(X+Y) = Var(X) + Var(Y) – Gono Nov 4 '16 at 10:49