Assume that $\mathbf{X}$ is a bivariate normal random variable $$\mathbf{\mu} = E\mathbf{X} = \begin{bmatrix} 0 \\ 2 \end{bmatrix} \ \text{and} \ \Sigma = Cov \ \mathbf{X} = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix} $$

What is the probability that $\mathbf{X}$ falls within the elipse corresponding to $(\mathbf{x} - \mathbf{\mu})^T \Sigma^{-1}(\mathbf{x} - \mathbf{\mu}) = 4.6 ?$

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So Im assuming I need to calculate the probability of $x_1$ lying between $x_1$ min and $x_1$ max, and likewise for $x_2$? However Im not sure how to do this calculation seeing as the variables $x_1$ and $x_2$ are not independent. It also seems like the endpoints of the ellipse do not match my calculations. I believe the ellipse should have half axe lengths $\sqrt{\lambda_i b}$ and its centered at $\mathbf{\mu}$. So according to my calculations $x_1$ should lie between -4.28 and +4.28 which seems incorrect from graphical inspection.

  • 1
    $\begingroup$ The random vector $\Sigma^{-1/2} (x-\mu)$ has distribution $N(0,I)$, so your example correspond to looking at probability that standard normal $N(0,I)$ is in the circle of radius $\sqrt{4.6}$ $\endgroup$
    – P. Quinton
    Commented Feb 12, 2020 at 14:04
  • $\begingroup$ How would you go about calculating the probability of the standard normal being inside the circle? Also Im assuming the circle is centered at $\mu$? $\endgroup$
    – Pame
    Commented Feb 12, 2020 at 14:46
  • $\begingroup$ Not sure what is being asked. Is it the probability $P\left[(\mathbf X-\mu)^T\Sigma^{-1}(\mathbf X-\mu)\le 4.6\right]$? Then you have $(\mathbf X-\mu)^T\Sigma^{-1}(\mathbf X-\mu)\sim \chi^2_2$ by a general result. $\endgroup$ Commented Feb 12, 2020 at 15:02

1 Answer 1


It is known that $U=\Sigma^{-1/2}(x-\mu)\sim \mathcal N(0,I)$ so you look at $\mathbb P(U\in \mathcal C)$ with $\mathcal C = \{ (x,y) : x^2+y^2\leq r^2 \}$. You can compute this by doing the change of coordinate from Cartesian to polar : \begin{align*} \int_{\mathcal C} dp_{U} &= \int_0^{2\pi} \int_0^r a \frac{1}{2\pi}\exp(-a^2/2) da d\theta\\ &= -\exp(-a^2/2) \bigg|_0^r\\ &= 1-\exp(-r^2/2) \end{align*} If you plug in $r=\sqrt{4.6}$ you get approximately $0.8997412$.


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