Given that $X_1, X_2, X_3 $ are independent random variables form $N(0, \sigma^2 )$, I have to indicate that the statistic given below has a $t$ distribution or not.
\begin{equation}
\frac{2X_1 – X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}
\end{equation}
In my attempt of solving this problem:
I start by showing that we can write the numerator as $a^TX$, where $a^T = (2 -1 -1)$ and $X^T= (X_1 X_2 X_3)$. Thus we have that $a^TX \sim N(0, a^T(\sigma^2 I)a)= N(0, 6\sigma^2)$. And so $\frac{1}{\sqrt{6\sigma^2}} a^TX \sim N(0,1)$ or $\frac{1}{\sqrt{6\sigma^2}}(2X_1-X_2-X_3)\sim N(0,1)$.
Next, we know that $(X_1+X_2+X_3) \sim N(0, 3\sigma^2)$. This implies that $\frac{1}{\sqrt{3\sigma^2}}(X_1+X_2+X_3) \sim N(0,1)$ and thus,$\frac{1}{{3\sigma^2}}(X_1+X_2+X_3)^2 \sim \chi^2(1)$. Similarly, $\frac{1}{2\sigma^2}(X_2-X_3)^2 \sim \chi^2(1)$. Therefore, $\frac{1}{{3\sigma^2}}(X_1+X_2+X_3)^2 + \frac{1}{2\sigma^2}(X_2-X_3)^2 \sim \chi^2(2)$ or $\frac{1}{{3\sigma^2}}\left((X_1+X_2+X_3)^2 + \frac{3}{2}(X_2-X_3)^2 \right) \sim \chi^2(2)$.
As a third step I have to show that $\frac{1}{\sqrt{6\sigma^2}}(2X_1-X_2-X_3)$ and $ \frac{1}{{3\sigma^2}}\left((X_1+X_2+X_3)^2 + \frac{3}{2}(X_2-X_3)^2 \right)$ are independent and I am not sure how to show that. Any help would be appreciated.
Best Answer
Consider the orthogonal transformation
$$\begin{pmatrix}Y_1 \\ Y_2 \\ Y_3\end{pmatrix}=\begin{pmatrix}\frac{2}{\sqrt 6} &-\frac1{\sqrt 6} & -\frac1{\sqrt 6} \\ \frac1{\sqrt 3} & \frac1{\sqrt 3} & \frac1{\sqrt 3}\\ 0 & \frac1{\sqrt 2} & -\frac1{\sqrt 2} \end{pmatrix}\begin{pmatrix}X_1 \\ X_2 \\ X_3\end{pmatrix}$$
So if $Y=(Y_1,Y_2,Y_3)^T$ and $X=(X_1,X_2,X_3)^T$, then $X\sim N(0,\sigma^2 I_3)\implies Y\sim N(0,\sigma^2 I_3)$.
Therefore,
$$T=\frac{2X_1 - X_2-X_3}{\sqrt{(X_1+X_2+X_3)^2 +\frac{3}{2} (X_2-X_3)^2}}=\frac{\sqrt 6Y_1}{\sqrt{3Y_2^2+3Y_3^2}}$$
I think you can take it from here.