Alg1231W12T3 - Continuous distributions

Sunday, 21 October 2018

12:06 AM

Machine generated alternative text: For x < 0 we have fn(x) This proves that fn (x) 2 0 Integrating the density over R gives For any natural number n • the function X for x > 0 cne¯Xdæ o otherwise is a probability density function We prove this by showing that fn(x) is not negative and it integrates to one O . And for x 2 0 we have [select all that apply] O for all x. fn (x)dc DO 1 where a O It can be shown (using integration by parts) that If we use this result; then we can conclude that as required fn (x)dx

 

Machine generated alternative text: Consider the Laplace distribution Since the exponential function is e ¯tdx 1 f@) 2 for —DO < < DO positive tor all real inputs O we can easily see that 2 f(x) dc O for all x e R. Integrating this density over R we get Because the density is an even function 1 e ¯lxldx. O , we can conclude that where a as required O We can caculate this as an improper integral, showing e¯Xdx = lim 2 —exp(-R) + 1

 

Machine generated alternative text: Let's calculate the mean and variance of the Gamma distributiom the probability density function of the Gamma distribution is defined to be X n —x for x > 0 o In general, the mean of a continuous random variable is defined to be So, in the case of the gamma distribution this is otherwise. xfn@) dc. oo 1 n+1e¯Xdx. Using a well know result (see note below) we can easily calculate this to be The variance can be calculate by Var(X) however it is much easier to calculate it using the alternative form Var(X) = E(X2) - E(X)2. The expectation of the X2 is given by n+2e¯Xdx. E(X2) — Use the same result as before (see note below) we can easily calculate this to be From this we calculate that Note: Recall that important formula for m E(X2) — Var(X) = E(X2) - n+l e

 

Machine generated alternative text: For a parameter a > 0 the probability density function of the Pareto distribution is defined to be fa(.c) — for x > 1 otherwise. O An interesting fact about the Pareto distribution is that the mean and variance is not defined for some choices of the parameter a. The mean of a continuous random variable with this distribution will be xfa@) dc Using the test from p-integrals from calculus, we know that this improper integral will only converge to a finite number when a > Note: you can type a in Maple notations by typing alpha. As before we'll calculate the variance using the formula Var(X) = E(X2) - E(X)2. The expectation of the X2 is given by E(X2) - Like before we know this will only converge to a finite number when a — 1 > 1 1 O a—I , or in other words, when a > 2 O O In this case we can find Calculating we get E(X2) ¯ alphai(alpfia-2) From this we can calculate that Var(X) (a—2)(a -1)2

 

Machine generated alternative text: -2 0.8 0.6 0.2 S Z 1) 0.341 2 z 3 The GeoGebra app above shows the probabilty density function for the standard normal distribution N(O, 1) _ The probability density function for this distribution is given by the expression 1 2K Answer the following questions about a random variable Z which has the standard normal distribution (i.e. Z i) P(—l < Z < 1) is about 68 ii) P(—2 < Z < 2) is about 95 iii) P(—1.5 < Z < 2.5) is about 93 iv)P(0.5 < Z < 1.5) is about 24 O O percent Note that this corresponds to the points within 1 percent Note that this corresponds to the points within 2 O percent O percent O standard deviation of the mean. O standard deviations of the mean

 

 

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