# Math problem in Pattern Recognition and Machine Learning

I had been carefully following Bishop's Pattern Recognition book and came across some of difficulties due to my merely basic math ground.First begin by the Gaussian section on page 80.

The functional dependence of the Gaussian on x is through the quadratic form Δ2 = (x − μ)TΣ-1(x − μ) (2.44) which appears in the exponent. The quantity Δ is called the Mahalanobis distance from μ to x and reduces to the Euclidean distance whenΣis the identity matrix. The Gaussian distribution will be constant on surfaces in x-space for which this quadratic form is constant.

What exactly is a constant quadratic form and distribution?How could (2.44) a constant.

And in page 82,

We now look at the moments of the Gaussian distribution and thereby provide an interpretation of the parameters μ and Σ. The expectation of x under the Gaussian distribution is given by

where we have changed variables using z = x − μ. We now note that the exponent is an even function of the components of z and, because the integrals over these are taken over the range (−∞,∞), the term in z in the factor (z + μ) will vanish by symmetry.Thus E[x] = μ (2.59)

And I could not see the magic from (2.58) to (2.59).

Another one regarding beta distribution.Why beta distribution has a probability more than 1 on its y-axis?

Sorry if all these sound stupid but it had been in my head for days.

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1) Since the multivariate Gaussian distribution has density $$\frac{1}{(2\pi)^{k/2}|\mathbf\Sigma|^{1/2}}\exp\left(-\frac{1}{2}({\mathbf x}-{\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x}-{\mathbf\mu})\right),$$
those $\mathbf x$ for which $({\mathbf x}-{\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x}-{\mathbf\mu})$ takes a given value will have the same density.
2) Letting $\mathbf z = {\mathbf x}-{\mathbf\mu}$ means the multivariate Gaussian distribution for $\mathbf z$ is symmetric about $\mathbf 0$ and so has an expectation of $\mathbf 0$. Using that, plus the integral of the whole density being $1$, allows you to see that the expectation of $\mathbf x$ is $\mathbf\mu$.
3) The beta distribution does not have a probability of more than $1$: it has a probability density more than $1$ in part of its support. Since it is restricted to taking values in the range $[0,1]$ and the integral of its density is $1$, this is inevitable (apart from the case of a uniform distribution).