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:

$$E(t) = 0$$


$$Var(t) = \frac{\vartheta }{\vartheta -2} = 2 $$

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?

  • 3
    $\begingroup$ A simple way to check if your answer is correct is to resort to one of the definition of t-distributed variables, $T = X / \sqrt{V/\nu}$, where $X$ is a standard normal and $V$ is a chi-squared with $\nu$ degrees of freedom. Does your calculated $C$ enable the fraction to match that? Hint: It should involve a $\sqrt{2}$. $\endgroup$
    – B.Liu
    Commented Aug 22, 2021 at 10:51
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    $\begingroup$ @Rob thank you for your comment. I will make the changes $\endgroup$
    – Kalvin
    Commented Aug 22, 2021 at 10:56
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    $\begingroup$ DBZrasengan16, help is available by clicking on the (?) (upper right) and by visiting this site's meta (extreme upper right hamburger icon). --- MathJax help: Short: quantumcomputing.meta.stackexchange.com/a/76/278 or long: math.meta.stackexchange.com/q/5020/510296 $\endgroup$
    – Rob
    Commented Aug 22, 2021 at 11:29
  • 1
    $\begingroup$ @Rob Thank you for your help. I have edited using your instructions. Is this fine? $\endgroup$
    – Kalvin
    Commented Aug 22, 2021 at 15:19
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    $\begingroup$ There are at least two fatal flaws in this approach: (1) your manipulations of the variance are incorrect and (2) you cannot hope to determine a distribution from its mean and variance alone. If you are focused on finding $C$ assuming the ratio is proportional to a Student t distribution, then computing a moment will indeed work. What exactly, then, is the problem you are trying to solve? Finding the distribution, finding $C,$ or learning how to compute with variances? $\endgroup$
    – whuber
    Commented Aug 22, 2021 at 15:19

1 Answer 1


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.


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 $C = \sqrt{\frac{3}{2}}$

  • $\begingroup$ $U$ is not a standard Normal variable. I suspect you mean to set $U = (X_1+X_2)/\sqrt{2}.$ $\endgroup$
    – whuber
    Commented Aug 23, 2021 at 16:40
  • $\begingroup$ @whuber yes i was thinking the same. I have made the edit. Let me know if it is correct? $\endgroup$
    – Kalvin
    Commented Aug 23, 2021 at 16:59
  • $\begingroup$ @Whuber, you're right $\endgroup$ Commented Aug 23, 2021 at 17:47

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