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The original poster ponders why, in a bivariate model, the SmoothKernelDistribution function ... using the Silverman 'rule of thumb' estimate of bandwidth ... yields results that do not match the original poster's calculations. I had the same issue in a univariate form. Short version It turns out that there are two versions of the Silverman rule of thumb ...

I'd like to add to my previous comment that the undocumented "Bandwidth" property exists for SmoothKernelDistribution (SKD) and KernelMixtureDistribution (MKD). Note that in the multivariate case there is a bit of inconsistency in how this is currently handled for SKD vs. KMD. SKD simplifies the bandwidth matrix when the off-diagonal elements are zeros. ...

Finally, I think I found it--or, at least, I found a formula "close enough" to the way Mathematica calculates the BW by default: hi = ((4*Subscript[\[Sigma], i]^5)/(3*n))^(1/5) where \[Sigma] is the standard deviation of the data. This should be performed to each component (e.g. i, j, k) of the dimensions. The reference came from the folowing link: ...

If $X\sim N\left(\mu ,\sigma ^2\right)$ and $Y=e^X$, then $Y\sim \text{Lognormal}(\mu ,\sigma )$. So, by selecting LogNormalDistribution[10, 10], you are effectively generating values from a $N(10, 100$) distribution (which is a very large variance), and then raising them to $e^X$ ... which will generate deliciously large variates. To see this: Here are 6 ...

f2[x_, mu2_, sigma2_] := 1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*Abs[(x - mu2)/sigma2]] Integrate[x*f2[x, mu, h], {x, -Infinity, Infinity}, Assumptions -> {Element[mu, Reals], h > 0}] (* mu *) Edit Let's make what he did wrong crystal clear to the OP. Should have used SetDelayed (:=) rather than Set (=) when defining f2. Needed to have an ...

One way to circumvent it is like this f2[x_, mu2_, sigma2_] := Piecewise[{{1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*((x - mu2)/sigma2)], x - mu2 >= 0}, {1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*((mu2 - x)/sigma2)], x - mu2 < 0}}] Integrate[x*f2[x, mu, sigma], {x, -Infinity, Infinity}, Assumptions -> sigma > ...

As I said in a comment, I'm not an expert on probability, but you can calculate the variance of a distribution with Variance: Variance[TransformedDistribution[ If[\[FormalX] == 1, 1 - 2 k, -1], \[FormalX] \[Distributed] BernoulliDistribution[p]]] // Simplify (* -4 (-1 + k)^2 (-1 + p) p *)