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In the answer for the following exercise:

Let $\{x_1,...,x_n\}$ be a finite collection of random variables with $E(x_i^2) \lt \infty$ ($i = 1,..., n$). Show that the set of all linear combinations $\Sigma_{i=1}^{n} \alpha_i x_i$ constitutes a vector space, which we denote by $L^\dagger_2$,

there's a statement that I don't really understand, that is:

The difficulty is that a random variable $y = \Sigma_{i}\alpha_i x_i$ might also be expressible as $y = \Sigma_{i}\beta_i x_i$, where $(\alpha_1,...,\alpha_n) \ne (\beta_1,...,\beta_n)$.

For simplicity, suppose that $n=2$ so we have only two random variables $x_1$ and $x_2$ in the collection. How could I construct an example where $y=\alpha_1 x_1 + \alpha_2 x_2=\beta_1 x_1 + \beta_2 x_2$ where $(\alpha_1, \alpha_2)\ne (\beta_1, \beta_2)$?

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    $\begingroup$ for your n=2 example, you would be able to do this exactly when $x_1 = a x_2$ for some constant $a$ (such as 1). In generally, you will be able to multiply express a linear combination in terms of its coefficients when the vectors constituting it are linearly dependent. $\endgroup$
    – John Madden
    Commented Dec 21, 2023 at 18:59
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    $\begingroup$ As emphasized by the context of the exercise, it isn't about random variables: it's purely about linear algebra. Let $x_2=-x_1,$ for instance. Then $\sum \alpha_i x_i = \alpha_1x_1+\alpha_2x_2=(\alpha_1-\alpha_2+\gamma)x_1+\gamma x_2$ for any real number $\gamma.$ $\endgroup$
    – whuber ♦
    Commented Dec 21, 2023 at 20:21
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    $\begingroup$ BTW, it's hard to see how that statement is any difficulty at all. All you need to do in this exercise is apply the closure property, which is that any linear combination of linear combinations of the $x_i$ is a linear combination of them. $\endgroup$
    – whuber ♦
    Commented Dec 21, 2023 at 20:23
  • $\begingroup$ @JohnMadden and whuber Thanks for your comments! It's simpler than I thought it was. $\endgroup$
    – Tran Khanh
    Commented Dec 22, 2023 at 2:47
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    $\begingroup$ @TranKhanh as I mentioned, there is no such case for N=2. Can you let me know what you don't like about my N=3 case? (note that in my example we have $0=x_3-\frac{x_1+x_2}{2}$). $\endgroup$
    – John Madden
    Commented Dec 23, 2023 at 14:54

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