A $2\times 2$, real, symmetric random matrix $M$ is provided with diagonal entries $M_{11}$ and $M_{22}$ following independent normal distributions $N(0,1)$ and off-diagonal entries $M_{12}=M_{21}$ following normal distribution $N(0,1/2)$.

What's the distribution and the mean of the absolute value of the difference between its two eigenvalues $\Delta E$ ?

Attempt :

The eigenvalues satisfy equation $x^2-Tx+\Delta=0$ where $T$ and $\Delta$ denote the trace and determinant of said matrix, respectively.

Therefore, $x_{\pm}=\frac 12(T\pm\sqrt{T^2-4\Delta})$ and $\Delta E=\sqrt{T^2-4\Delta}=\sqrt{(A-D)^2+4BC}$

($A=M_{11}, B=M_{12}, C=M_{21}, D=M_{22})$

$A$ and $D$ are two independent Gaussian variables with mean $0$ and variance $1$, therefore $A-D$ is a Gaussian random variable with mean $0$ and variance $2$, $N(0,2)$.

$B=C$ is a Gaussian random variable with mean $0$ and variance $1/2$

Therefore, $4BC=\frac1416B^2=\frac14(4B)^2$ with $4B \sim4N(0,1/2)=N(0,2)$

So, $\Delta E \sim\sqrt{N(0,2)^2+\frac14N(0,2)^2}=\sqrt{\frac54 N(0,2)^2}=\sqrt{N(0,\sqrt{5})^2}=|N(0,\sqrt{5})|$

This happens to be a folded normal distribution with $\mu=0$ and $\sigma=\sqrt{\sqrt{5}}$

So the mean should be $\sqrt{\frac{2\sqrt{5}}{\pi}}\approx 1.1931$.

However, the following MATLAB code seems to suggest otherwise :

for i=1:1e+5

returns $1.5241$

Am I handling distributions wrongly again ?

Context : lonely physics student who wanted to learn maths, uni says it's useless for a physicist and that adults learn by themselves, gives (ungraded) exercises to train instead, but I can't do a single one correctly, despite spending a whole day on the first question of a single problem. Can't improve no matter what. Work hard and you'll get there, they said. $\text{*roll eyes*}$

  • $\begingroup$ At some point in your calculation you're failing to keep track of independence. I think it's when you add $N(0, 2)^2$ to $\frac{1}{4} N(0, 2)^2$. These are independent, so their sum isn't $\frac{5}{4} N(0, 2)^2$. $\endgroup$ Feb 25, 2017 at 21:27
  • $\begingroup$ @Qiaochu Thanks, I guess you're right. I'll investigate this $\endgroup$
    – Evariste
    Feb 25, 2017 at 21:36
  • $\begingroup$ @Qiaochu Is this problem solvable ? It seems like the weighted sum of independent chi squared doesn't correspond to any elementary distribution, and the convolution trick for sums of independent variables doesn't yield anything... no closed form, at least $\endgroup$
    – Evariste
    Feb 25, 2017 at 23:11
  • $\begingroup$ @Evariste, you use second parameter in $N(0,1/2)$ for variance or for standard deviation? In Matlab code above it is SD. When you find distribution of $4B$, it al SD also. And when you use $N(0,2)$, here $2$ is the variance? $\endgroup$
    – NCh
    Feb 26, 2017 at 4:01
  • $\begingroup$ @NCh The second parameter was meant to be the variance. Thanks for pointing out! Now Matlab gives a mean roughly equal to $\sqrt{\pi}\approx 1.7721$ $\endgroup$
    – Evariste
    Feb 26, 2017 at 9:11

1 Answer 1


I presume that $M_{12}$ is a Gaussian random variable with mean $0$ and variance $1/2$ (not SD!) and is independent on $M_{11}$ and $M_{22}$. Then $$\Delta E=\sqrt{T^2-4\Delta}=\sqrt{(A-D)^2+4BC}=\sqrt{X^2+Y^2},$$ where $X=(A-D)$ is also a Gaussian random variable with mean $0$ and variance $2$, $N(0,2)$, and $Y=2M_{12}$ is a Gaussian random variable with mean $0$ and variance $2^2\cdot \frac12=2$. Since $X$ and $Y$ are independent, $\dfrac{X}{\sqrt2}$ and $\dfrac{Y}{\sqrt2}$ are standard Gaussian, then $$ \dfrac{X^2+Y^2}{2} \sim \chi^2_2=Exp(\frac12), $$ where $\chi^2_2$ is the chi-squared distribution with $2$ degrees of freedom. This distribution is exactly the same as the Exponential distribution with mean $2$.

Then $$\Delta E=\sqrt{X^2+Y^2}=\sqrt{2Exp(\frac12)}=\sqrt{Exp(\frac14)}.$$

If you need mean value, calculate $$E(\Delta E)=\int_0^\infty \sqrt{x} \frac14 e^{-\frac14x}\,dx=\sqrt{\pi}\approx 1.77245.$$

If you ask M(1,2)=normrnd(0,1/2) in Matlab, you generate $N(0,1/4)$. To generate $N(0,1/4)$, write M(1,2)=normrnd(0,1/sqrt(2))and compare results.

To find distribution of $\Delta E=\sqrt{Exp(\frac14)}$ write CDF for $y>0$: $$F_{\Delta E}(y)=P(\sqrt{Exp(\frac14)}\leq y)=P(Exp(\frac14)\leq y^2)=1-e^{-\frac{y^2}{4}}. $$ The probability density function is $$ f_{\Delta E}(y)=\frac{y}{2}e^{-\frac{y^2}{4}},\quad y>0. $$


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