Finding mean and variance of t-distribution to solve for constant c Suppose that 5 Random Variable X1, X2,... X5 are independent and each has standard normal distribution. A constant C such that the random variable
$$C\frac{X_{1} + X_{2}}{(X_{3}^2 + X_{4}^2 + X_{5}^2) ^ {\frac{1}{2}}}$$
will have a t-distribution?
I began to solve this by taking the mean and variance of the above random variable(lets call this RT).
Given the following:
$$E(t) = 0$$
and
$$Var(t) = \frac{\vartheta }{\vartheta -2} = 2 $$
Taking the variance of the random distribution RT
$$Var(RT) = C^{2}\frac{(var(X_{1})+ var(X_{2}))}{(var(X_{3}^2)+ var(X_{4}^2+ var(X_{5}^2))^{\frac{1}{2}}}$$
which is equal to
$$Var(RT) = C^{2}\frac{1+ 1}{(3+3+3 )^{\frac{1}{2}}}\ as \ Var(X^2)=3\sigma^2 \ and\ Var(X)=1$$
Therefore equating the above to 2 gives
$$C^{2} = 3\ and\ C = 3^{1/2}$$
Is my answer correct?

In case you want to find $C$ such that $$C\frac{X_1+X_2}{\left(X_3^2+X_4^2+X_5^2\right)^{1/2}}$$ follow a $t$-distibution. Here is how you should proceed.
Answer
As $X_1$ and $X_2$ are independents and standard normal distributed, $X_1+X_2\sim \mathcal{N}(0,2)$ and then $U := \frac{1}{\sqrt{2}}(X_1+X_2)$ is a standard normal random variable.
As $X_3, X_4$ and $X_5$ have standard normal distribution, $V := X_3^2+X_4^2+X_5^2$ has a $\mathcal{X}^2$ distribution with degree of freedom $\nu=3$.
After all, we know that $\frac{U}{\sqrt{V/\nu}}=\frac{\frac{1}{\sqrt{2}}(X_1+X_2)}{\sqrt{\left(X_3^2+X_4^2+X_5^2\right)/3}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{3}}}\frac{X_1+X_2}{\left(X_3^2+X_4^2+X_5^2\right)^{1/2}}$ follow a $t$-distribution.
So $C = \sqrt{\frac{3}{2}}$
