fx n y fx y the previous equation can be rewritten

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Unformatted text preview: , we can rewrite the previous equation as, n FZ z 1 1 FX i z 1 1 FX z n i 1 So, by differentiating, we obtain the pdf as, f Z z d n 1 FZ z n1 FX z f X z dz Problem 2. The pdf of peak flow of flood events at a gauging station is exponentially distributed as, f X x e x for x 0 with 0.3 . If five floods are observed in a year, determine the probability distribution of the maximum peak flow of flood occurrences. Solution. Given that, f X x e x for x 0 Therefore by integrating with respect to x , we get, y y FX y f X x dx e x dx 0 0 1 e y for y 0 Assuming that the peak flows of the flood occurrence are independent of each other, we get, FY y FX y 1 e 0.3 y n 5 for y0 By differentiating we obtain the pdf (i.e. maximum intensity) as, fY y nFX y n 1 f X y 4 5 1 e 0.3 y 0.3 e 0.3 y 4 1.5 1 e 0.3 y e 0.3 y for y 0 Similarly, the minimum intensity can also be found out as, FY y FX y 1 e 0.3 y n 5 for y0 Here, it should be noted that, this expression follows the same distribution of peak flood flow, X. 5. Product and Quotient of Two Random Variables 5.1. Product of two random variables Let X and Y are two random variables. Now, the probability distribution of another variable which is the product of the two i.e. Z X Y can be determined by integrating the joint pdf of X and Y for those values of the product X Y that are less than or equal to z...
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