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.
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 $\Sigma_{ii}$: $$x_i\sim\mathcal N(\mu_i,\Sigma_{ii})$$
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.
