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A rather elementary question: If we have:

$X = X_1 + X_2 + ... + X_n$ and $X_i \sim N(0,\sigma^2)$ and independent,

can we express $X$ as $\sqrt n X_i$ for any $i = 1,...,n$ ?

I think since all the components $X_i$ follow the normal distribution and we are summing $n$ of them up, the resulting rv is $X \sim N(0, n\sigma^2)$ and therefor we can use $X = \sqrt n X_i$.

Is this reasoning correct? If yes, under what condition it can be false?

x0dros
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2 Answers2

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A random variable is a real-valued function, $X_j = h_j(y),\; y\in S_y$, where $S_y$ is the domain of the function. A function is fully determined by three things: its domain, its range, and its functional form. So let's check: Although it is not mentioned, let's say that the $X$ variable and the $\sqrt n X_i$ r.v. have the same domain, say, the reals (that domain has nothing to do with the distribution they follow). Since they are both normally distributed, they have the same range (which is the "support" i.e. it is the domain of the distribution they follow), $(-\infty, \infty)$. Do they have the same functional form? Unlikely. We have $X = \sum_{j=1}^nX_j = \sum_{i=1}^nh_j(y)$ while $X_i = \sqrt nh_i(y)$. For these to be identical we should have $$\frac {1}{\sqrt n -1}\sum_{j\neq i}^nh_j(y) = h_i(y),\;\; \forall y$$ This relation may, or may not hold. So in general, writing $X=\sqrt n X_i$ is wrong.

But, they are identically distributed... what does that mean? A function becomes a random variable if its range has such properties which can be summarized by another function possessing certain properties, a "cumulative distribution function", and in our case by the latter's derivative, the probability density function also (since it exists). But we just saw that two random variables may have identical distributions and at the same time they may be different functions. To put it in symbols, for two r.v.'s, $$Z = Y \Rightarrow Z\sim_d Y$$ (and so also $E(Z^r) = E(Y^r) \;\; \forall r$), but

$$Z\sim_d Y \nRightarrow Z = Y$$

Yet another way to think of it intuitively is the following: assume that $X_i$ acquires a specific value, say $x_i^*$. Is it true that then the other random variables will acquire such values so as we will necessarily have $\sum_{j\neq}^nx^*_j = (\sqrt n-1) x_i^*$? No. Therefore, $X$ and $\sqrt nX_i$ are not identical functions.

See also Can two random variables have the same distribution, yet be almost surely different?

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Alecos Papadopoulos
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  • The reason I asked the OP is because in another question I posted [here](http://stats.stackexchange.com/questions/88628/distribution-of-the-ratio-of-dependent-chi-square-random-variables) I was left with the impression that I can do it. Your reasoning here is undoubtedly correct, so I assume either I did not understand the answer given there correctly, or the commentator was wrong. – x0dros Mar 06 '14 at 02:29
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    I read the comments, and my impression that they all refer to the distributional relations between random variables. We usually say "identical random variables" but we only mean "identically distributed" not "identical as functions". – Alecos Papadopoulos Mar 06 '14 at 02:41
  • Is it correct then to say $X \sim \sqrt n X_i \sim N(0,n\sigma^2)$ ? – x0dros Mar 06 '14 at 14:39
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    Yes it is, this notation expresses the "identically distributed" aspect. – Alecos Papadopoulos Mar 06 '14 at 15:17
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$Cov(X,\sqrt{n}X_i)=\sqrt{n}\sigma^2$, but if $X$ were $\sqrt{n}X_i$, then we'd have $Cov(X,\sqrt{n}X_i)=n\sigma^2$

hence, the answer is No

Aksakal
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