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Let $X$ be a vector-valued random variable with variance $\mathbb{V}[X] < \infty$. How is the variance of $X$ along a unit-vector $\hat{v}$ defined? Can we say that in general it is $\hat{v}^\top \mathbb{V}[X]\hat{v}$?

Example: For $X\sim \mathcal{N}(\mu, \Sigma)$ with $\mathbb{V}[X] = \Sigma$ then the variance along the unit vector is $\mathbb{V}_{\hat{v}}[X] = \hat{v}^\top \Sigma \hat{v}$.

Euler_Salter
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  • How is the variance of *any* vector defined? – whuber Jan 24 '22 at 16:55
  • @whuber I'd say the variance of a vector-valued random variable is $\mathbb{V}[x] = \mathbb{E}[(x-\mathbb{E}[x]) (x - \mathbb{E}[x])^\top]$ based on [this](https://en.wikipedia.org/wiki/Variance#For_vector-valued_random_variables) – Euler_Salter Jan 24 '22 at 16:56
  • What exactly do you mean by the variance along a unit vector? Do you mean $\mathbb{V}\bigg[ \vec v^TX\vec v \bigg]?$ – Dave Jan 24 '22 at 17:01
  • @Dave I am trying to understand how to write it formally, so I am not sure yet. Essentially I would like to see the variance of that random variable when projected in that direction, I suppose – Euler_Salter Jan 24 '22 at 17:03
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    You would have to have a multivariate random variable to project it, right? – Dave Jan 24 '22 at 17:04
  • @Dave Yes absolutely, it's a vector-valued random variable. I will make it clear in the question – Euler_Salter Jan 24 '22 at 17:05
  • Immagine you have a diagonal 2D Gaussian with variances $1.0$ and $5.0$ respectively. Then the variance along $(1,0)$ is just $1.0$ and along $(0,1)$ is $5.0$. How does this generalise to non gaussian and larger dimensional problems? – Euler_Salter Jan 24 '22 at 17:16
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    When you apply your definition to the random variable $v^\prime X,$ you immediately obtain the formula in terms of $\Sigma:$ this is the *bilinearity* property of variance, following directly from the linearity of expectation. See https://stats.stackexchange.com/questions/184998 for an elementary (matrix-free) exposition. – whuber Jan 24 '22 at 17:18

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