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Unformatted text preview: ECEN 303: Assignment 7 Problems: 1. Let X have the PDF f X ( x ) = 2 e- | x | , where is a positive scalar. Verify that f X satisfies the normalization condition, and evaluate the mean and variance of X . Consider the integral integraldisplay - 2 e- | x | dx = integraldisplay- 2 e x dx + integraldisplay 2 e- x dx = integraldisplay e- x dx = 1 . Thus, this PDF satisfies the normalization condition. The mean of this random variable is E[ X ] integraldisplay - x 2 e- | x | dx = integraldisplay- x 2 e x dx + integraldisplay x 2 e- x dx = 0 . The mean can also be obtained by inspection, noting that the integrant is an odd function. Finally, the variance of X is Var( X ) = E[ X 2 ] = integraldisplay - x 2 2 e- | x | dx = integraldisplay- x 2 2 e x dx + integraldisplay x 2 2 e- x dx = integraldisplay x 2 e- x dx = bracketleftbigg- x 2 e- x + 2 x e- x- 2 2 e- x bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle = 2 2 . 2. Consider a traingle and a point chosen within the triangle according to the uniform probability law. Let X be the distance from the point to the base of the triangle. Given the height of the triangle, find the CDF and the PDF of X . Denote the height of the triangle by h , and the width of the base by b . For x (0 , h ), the CDF is given by F X ( x ) = Pr( X x ) = x 2 parenleftBig b + b ( h- x ) h parenrightBig bh 2 = x parenleftBig 1 + ( h- x ) h parenrightBig h = x h parenleftBig 2- x h parenrightBig . The PDF can be obtained by differentiating the CDF, f X ( x ) = d dx x h parenleftBig 2- x h parenrightBig = 2 h- 2 x h 2 ....
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- Fall '07