কোর্স:পারিসাংখ্যিক সংকেত প্রক্রিয়ণ/দৈব ভেক্টর ও প্রক্রিয়া
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প্রশিক্ষক: খান মুহাম্মদ বিন আসাদ |
Let <math>\mathbf{v}</math> be a random vector. It's CDF is defined as,
- <math>F(v_0) = \text{Pr}(\mathbf{v} \le v_0)</math>.
The right-top plot shows a CDF for a bivariate normal distribution with <math>\mu=[1, -1]</math> and <math>\sigma=[0.9, 0.9]</math>. Characteristics of a CDF as obvious in the plot are:
- <math>F(\infty) = 0</math>, <math>F(\infty) = 1</math>
- if <math>a \le b</math>, <math>F(a) \le F(b)</math>
- <math>
\begin{align} p(x) &= \lim\limits_{\Delta x\rightarrow 0} \frac{\text{Pr} (x<X<x+\Delta x)}{\Delta x} \\ &= \lim\limits_{\Delta x\rightarrow 0} \frac{F(x+\Delta x) - F(x)}{\Delta x} \end{align} </math>
characteristics of PDF are,
- <math>p(x) \ge 0</math>
- <math>\int\limits_{-\infty}^\infty p(x) dx = 1</math>
PDF for the same bivariate normal distribution is shown on the right.
- PDF, <math>p(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}</math>
- CDF, <math>f(x) = \frac{1}{2}\left[1+\text{erf}\left(\frac{x-\mu}{\sqrt{2\sigma^2}}\right)\right]</math>
- <math>p(x,y) = \frac{1}{\sqrt{2\pi\sigma_1^2}} \frac{1}{\sqrt{2\pi\sigma_2^2}} e^{-\frac{(x-\mu_1)^2}{2\sigma_1^2}} e^{-\frac{(x-\mu_2)^2}{2\sigma_2^2}}</math>
- <math>f(x,y) = \frac{1}{4}\left[1+\text{erf}\left(\frac{x-\mu_1}{\sqrt{2\sigma_1^2}}\right)\right] \left[1+\text{erf}\left(\frac{x-\mu_2}{\sqrt{2\sigma_2^2}}\right)\right]</math>
- <math>p(x,y) \propto \exp(-\frac{x^2}{2\sigma_1^2} - \frac{xy}{\sigma_{12}^2} - \frac{y^2}{2\sigma_2^2})</math>
- marginal PDF, <math>p(\mathbf{x}) = \int\limits_{-\infty}^\infty p(x,y) \ dy</math>
Transformation
x is a continuous random variable with PDF p(x), and <math>y=g(x)</math> is a single valued differentiable function.
<math>p(y) = p(x) \left\vert \frac{dx}{dy} \right\vert = p(g^{-1}(y)) \left\vert \frac{d}{dy} g^{-1}(y) \right\vert</math>
The differentiation term is called the Jacobian, <math>J(y,x)</math>, of the transformation.
e.g. x is drawn from <math>\mathcal{N}(0,\sigma^2)</math>, <math>y=x^2</math>
<math> \begin{align} p_y(y) &= 2.\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{x^2}{2\sigma^2}} \frac{1}{2\sqrt{y}} \\ &= \frac{1}{\sqrt{2\pi\sigma^2}} \frac{e^{-y/(2\sigma^2)}}{\sqrt{y}} \end{align} </math>
<math>p_y(y)</math> is the chi-squared distribution with 1 degree of freedom, <math>\chi^2(y,1)</math>
This can be generalized for vectors,
<math>\mathbf{y} = \mathbf{g}(\mathbf{x})</math>
Volume, <math>V_x = \Delta x_1 \Delta x_2 \Delta x_3 ... \Delta x_N</math>
<math>V_y = J(\mathbf{y}, \mathbf{x}) V_{\mathbf{x}}</math>
<math>J(\mathbf{y}, \mathbf{x}) = \begin{vmatrix} \frac{\partial g_1(\mathbf{x}) }{\partial x_1 } & \frac{\partial g_2(\mathbf{x}) }{\partial x_1 } & \cdots & \frac{\partial g_N(\mathbf{x}) }{\partial x_1 } \\ \frac{\partial g_1(\mathbf{x}) }{\partial x_2 } & \frac{\partial g_2(\mathbf{x}) }{\partial x_2 } & \cdots & \frac{\partial g_N(\mathbf{x}) }{\partial x_2 } \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial g_1(\mathbf{x}) }{\partial x_N } & \frac{\partial g_2(\mathbf{x}) }{\partial x_N } & \cdots & \frac{\partial g_N(\mathbf{x}) }{\partial x_N } \end{vmatrix} </math>
<math>p_y(\mathbf{y}) = \frac{1}{J(\mathbf{y}, \mathbf{x})} p_x(\mathbf{g}^{-1}(\mathbf{y}))</math>