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We draw independently six numbers ($x_1,x_2,x_3,x_4,x_4,x_5,x_6$) from a same distribution with expected value $m$ and variance $\sigma ^2$.

Next, we create a matrix

$\begin{bmatrix} x_1&x_2&x_3\\x_2&x_4&x_5\\x_3&x_5&x_6\end{bmatrix}$

I have to calculate the expected value of its determinant as a function of $m$ and $\sigma^2$.

So, I don't have idea how I can start.

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closed as off-topic by Did, José Carlos Santos, mrtaurho, YiFan, Leucippus Feb 3 at 0:44

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    $\begingroup$ "Idea how you can start": Expand the determinant as an alternated sum of six products of the $x_k$s and compute the expectation of each product. $\endgroup$ – Did Feb 1 at 22:49
  • $\begingroup$ I have taken the liberty to modify your title and a little your text in order that more pople can beneficiate from the problem and its results. Besides, please note that I have added to my answer a simulation histogram with comments. $\endgroup$ – Jean Marie Feb 2 at 0:24
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Begin by expand your determinant using the rule of Sarrus (https://en.wikipedia.org/wiki/Rule_of_Sarrus)

$$D=\det \begin{bmatrix} x_1&x_2&x_3\\x_2&x_4&x_5\\x_3&x_5&x_6\end{bmatrix}$$

$$=x_1x_4x_6+2x_2x_3x_5-x_4x_3^2-x_1x_5^2-x_6x_2^2\tag{1}$$

Due to independence,

$$\text{if} \ p \neq q : \ \ \ E(x_px_q)=E(x_p)E(x_q)\tag{2}$$

Because operator $E()$ is linear, we can say that the expected value of (1) is :

$$E(D)=E(x_1)E(x_4)E(x_6)+2E(x_2)E(x_3)E(x_5)-E(x_4)E(x_3^2)-E(x_1)E(x_5^2)-E(x_6)E(x_2^2)$$

$$=m^3+2m^3-mE(x_3^2)-mE(x_5^2)-mE(x_2^2).\tag{3}$$

Now you must remember that (2) is not true if $p=q$. We must use the fact that :

$$E(x_p^2)=E(x_p)^2+\sigma^2=m^2+\sigma^2\tag{4},$$ and it remains to replace (4) into (2) to obtain :

$$E(D)=-3m\sigma^2.$$

I have simulated the case where all the $x_p$s are drawn from a uniform distribution $U[0,1]$, with mean $m=\frac12$ and variance $\sigma^2=\tfrac{1}{12}$ (https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)). The experimental variance one obtains is very close to $-\tfrac{1}{8} $ with a funny witch's hat histogram giving a good idea of the underlying pdf with a left trail a little heavier than its right counterpart, explaining the (slightly) negative mean :

enter image description here

It is not uninteresting to see that the extreme values for $D$ look to be in this case $-2$ and $2$ (I have no proof), attained with the following determinants:

$$D=\det \begin{bmatrix} 1&1&0\\1&0&1\\0&1&1\end{bmatrix}=-2 \ \ \text{and} \ \ D=\det \begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}=2.$$

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Explicitly the determinant $=x_1(x_4x_6-x_5^2)-x_2(x_2x_6-x_3x_5)+x_3(x_2x_5-x_3x_4)$. Thee are three (positive) terms where all $x's$ are different, so the expectations (independence) are $m^3$, while there are three (negative) terms containing one square, so the expectations are $(\sigma^2+m^2)m$. Combining these to get the expectation of the determinant to be $-3m\sigma^2$.

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    $\begingroup$ You are correct. Typo (capital +, lower case = on key). $\endgroup$ – herb steinberg Feb 2 at 2:31

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