By Kaddour Najim, Enso Ikonen, Ait-Kadi Daoud
For engineers facing stochastic tactics and for college kids of computerized regulate and mechanical and electric engineering, Najim (INP, Toulouse, France) Enso Ikonen (U. of Oulu, Finland) and Ait-Kadi Daoud (mechanical engineering, U. of Laval, Quebec) give some thought to in flip stochastic tactics, estimation, optimization, and the research of recursive stochastic algorithms to provide an explanation for the way to process a number of difficulties in utilized chance and data. They emphasize the layout of techniq ...
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33 The joint density function of X, Y is f (x, y) = 1 −(y+x/y) e , 0 < x, y < ∞ y (a) Verify that the preceding is a joint density function. (b) Find Cov (X, Y ). Solution: To show that f (x, y) is a joint density function we need to show it is ∞ ∞ nonnegative, which is immediate, and that −∞ −∞ f (x, y) d y d x = 1. 21) that E[Y ] = 1 We compute E[X ] and E[X Y ] as follows: E[X ] = = ∞ ∞ x f (x, y) d y d x −∞ −∞ ∞ ∞ −y e 0 0 x −x/y e dx dy y ∞ Now, 0 xy e−x/y d x is the expected value of an exponential random variable with parameter 1/y, and thus is equal to y.
Suppose first that X and Y are continuous, X having probability density f and Y having probability density g. 17) The cumulative distribution function FX +Y is called the convolution of the distributions FX and FY (the cumulative distribution functions of X and Y , respectively). 36 (Sum of Two Independent Uniform Random Variables) If X and Y are independent random variables both uniformly distributed on (0, 1), then calculate the probability density of X + Y . 18), since 1, 0 < a < 1 0, otherwise f (a) = g(a) = we obtain f X +Y (a) = 1 f (a − y) dy 0 For 0 ≤ a ≤ 1, this yields f X +Y (a) = a dy = a 0 For 1 < a < 2, we get f X +Y (a) = 1 dy = 2 − a a−1 Hence, ⎧ 0≤a≤1 ⎨a, f X +Y (a) = 2 − a, 1 < a < 2 ⎩ 0, otherwise Rather than deriving a general expression for the distribution of X +Y in the discrete case, we shall consider an example.
11 If the number of accidents occurring on a highway each day is a Poisson random variable with parameter λ = 3, what is the probability that no accidents occur today? 12 Consider an experiment that consists of counting the number of α-particles given off in a one-second interval by one gram of radioactive material. 2 such α-particles are given off, what is a good approximation to the probability that no more than two α-particles will appear? 2. 382 2 Continuous Random Variables In this section, we shall concern ourselves with random variables whose set of possible values is uncountable.