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I am faced with the above question, I'm aware that a linear combination of a normally distributed variable maintains its normality, however in this question it does not seem to be facing a linear combination. Nor did it first resemble a chi-squared to me.

That said my original (and probably very incorrect) working went something like:

$$Y=\sum_{i=1}^{N}\sigma^{-2}(X_i-\mu)^2$$ $$=\sigma^{-2}\sum_{i=1}^{N}(X_i-\mu)^2$$

As $$\sigma^2=\frac{1}{N-1}\sum_{i=1}^{N}(X_i-\mu)^2$$ then $$\sum_{i=1}^{N}(X_i-\mu)^2=(N-1)\sigma^2$$

Therefore $$ Y=\sigma^{-2}(N-1)\sigma^2 $$

$$Y=N-1$$ which is obviously just a constant, and I imagine completely wrong. Intuitively Y can't be negative, and therefore can't be normally distributed around 0, unless it's variance was 0 (though I'm not sure that would still be considered a normal distribution?).

And therefore my best guess would be it's (c) - which I can sort of understand, $\sum_{i=1}^{N}\frac{(X_i-\mu)^2}{\sigma^2}$ effectively acts as a way to standardise the distribution, and it's then squared and summed, giving a Chi-squared with N degrees of freedom.

Any guidance on this would be greatly appreciated.

  • 1
    $\begingroup$ Please add the self-study tag and read its tag-wiki. In approaching this, first consider what happens for $N=1$ $\endgroup$
    – Glen_b
    Sep 21, 2015 at 16:00

1 Answer 1



$$ Z_i = \frac{X_i - \mu}{\sigma}. $$

What is the distribution of $Z_i$ (be specific)? Can you write $Y$ in terms of the $Z_i$ instead of the $X_i$?

  • $\begingroup$ $Z_i$ is $\mathcal{N}(0,1)$ $Y=\sum^n_{i=1} Z^2_i$ Therefore $Y$ is $\chi_N$ gotcha $\endgroup$
    – Nik-D
    Sep 21, 2015 at 17:40
  • $\begingroup$ Ah yes, sorry relatively new to the forum! $\endgroup$
    – Nik-D
    Sep 21, 2015 at 17:48
  • $\begingroup$ Thanks. As per your math, the first equation with $N-1$ is an estimate of standard deviation, as opposed to $\sigma$ which is the standard deviation. $\endgroup$ Sep 21, 2015 at 18:07

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