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I want to find the univariate normals of a multivariate normal in order to plot them. If we assume that each are independent, then I know that we can use the mean vector and diagonal of the covariance of the multivariate normal can be used as the means and variances of those independent univariate normal distributions.

Is it same to assume so or are there any other technique to derive the univariate normals of a multivariate normal?

I appreciate any help resolving the above doubt.

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  • $\begingroup$ Are you talking about determining all the 1-dimensional marginals? If so you can get them by integrating out all the other components from the joint distribution. $\endgroup$ – Michael Chernick Apr 17 '18 at 6:27
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    $\begingroup$ @MichaelChernick, I guess so. According to the wikipedia the marginal distribution of multivariate normal distribution is derived by dropping the irrelevant variables. Therefore, can't I just used the individual mean and variance of each variable to find its 1-dimensional marginal? $\endgroup$ – Nadheesh Apr 17 '18 at 7:49
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If you have a multivariate normal with means $\mu$ and covariance matrix $\Sigma$, then the univariates (marginals) are simple normals with corresponding mean $\mu_i$ and the variance $\Sigma_{ii}$: $$x_i\sim\mathcal N(\mu_i,\Sigma_{ii})$$

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Your comment:

Therefore, can't I just used the individual mean and variance of each variable to find its 1-dimensional marginal?

This is exactly what you can do. To see this in the case for a bivariate normal $$\left(\begin{array}{c}X_1 \\X_2 \end{array}\right) \sim N_2\left(\left(\begin{array}{c}\mu_1 \\\mu_2 \end{array}\right), \left(\begin{array}{cc}\sigma^2_1 & \sigma_{12} \\ \sigma_{12} & \sigma^2_2 \end{array}\right) \right)\,. $$

Let $\rho = \sigma_{12}/\sigma_{1} \sigma_2$. The pdf of this random variable is

$$ \Tiny f(x_1, x_2) = \dfrac{1}{2 \pi \sigma_2 \sigma_2 \sqrt{1 - \rho^2}} \exp \left\{-\dfrac{1}{2\sqrt{1-\rho^2}}\left( \dfrac{(x_1 - \mu_1)^2}{\sigma^2_1} - \dfrac{2\rho(x_1 - \mu_1)(x_2 - \mu_2)}{\sigma_1 \sigma_2} + \dfrac{(x_2 - \mu_2)^2}{\sigma_2^2} \right)\right\} \,.$$

To obtain the marginal distribution of $X_1$, we integrate out $X_2$. You can find a proof of that here. For the multivariate normal case, you can find a proof here.

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    $\begingroup$ +1 -- but the proof is dead simple. It amounts to observing that after integrating out all the other variables, the integrand is still the exponential of a quadratic form in the remaining variable and that its leading term and linear term do not change. (That is, there's no need actually to perform the integration or even to write out the PDF explicitly.) Those facts uniquely determine the answer. $\endgroup$ – whuber Apr 17 '18 at 15:51

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