Ex 1.8 - Show that the mean value of Gaussian distribution is µ. (Pattern Recognition and ML by Christopher Bishop)

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Using ChatGPT by the formula for E[x,y] when x and y are discrete

If xn and xm are values from a gaussian distribution, can you derive the expression for E[xn.xm]

If $X_n$ and $X_m$ are values from independent Gaussian distributions, each with its own mean ($\mu_n$ and $\mu_m$) and variance ($\sigma_n^2$ and $\sigma_m^2$), and you want to find the expected value of their product $E[X_n \cdot X_m]$, it is important to note that the product of independent Gaussian variables does not follow a Gaussian distribution.

However, if you are interested in the covariance of $X_n$ and $X_m$, you can use the fact that the covariance of independent variables is zero:

$Cov[X_n, X_m] = E[X_n \cdot X_m] - E[X_n] \cdot E[X_m]$

Since $X_n$ and $X_m$ are independent, $Cov[X_n, X_m] = 0$, and you get:

$E[X_n \cdot X_m] = E[X_n] \cdot E[X_m]$

So, the expected value of the product of independent Gaussian variables is the product of their means:

$E[X_n \cdot X_m] = \mu_n \cdot \mu_m$

No, no. xn and xm are from the same distribution

I apologize for the confusion. If $X_n$ and $X_m$ are from the same Gaussian distribution, let's denote their mean as $\mu$ and their variance as $\sigma^2$. In this case, you're asking for the expected value $E[X_n \cdot X_m]$.

Since $X_n$ and $X_m$ are from the same distribution, we can simplify it as follows:

$E[X_n \cdot X_m] = E[X_n]^2 + Cov[X_n, X_m]$

Now, since $X_n$ and $X_m$ are from the same distribution, their covariance simplifies to the variance:

$E[X_n \cdot X_m] = \mu^2 + \sigma^2$

So, the expression for $E[X_n \cdot X_m]$ for two variables from the same Gaussian distribution is the square of the mean plus the variance of that distribution.