How to solve these four questions of continuous distributions?

1.

For any natural number n = 0, 1, 2, . . . the function fn(x) = n!" re - for 1 2 0 0 otherwise is a probability density function. We prove this by showing that fn (x) is not negative and it integrates to one. For x 0 we have [select all that apply] Deco Ve >O. This proves that fn (I) > for all c. Integrating the density over R gives In(z)dx = where a = + It can be shown (using integration by parts) that re "de = n!. If we use this result, then we can conclude that In(z)do = + as required.Consider the Laplace distribution f(z) = for -00 (X for all c c R. Integrating this density over R we get f(x) da = / Le kidz. Because the density is an even fu... , we can conclude that f(z)da = 2 / e da = e -I de where a= X We can caculate this as an improper integral, showing R lim e "de = lim R-+00 JO R-+DO as required.For a parameter o > 0 the probability density function of the Pareto distribution is defined to be fa(I) = za+1 for 1 2 1 0 otherwise. An interesting fact about the Pareto distribution is that the mean and variance is not defined for some choices of the parameter or. The mean of a continuous random variable with this distribution will be E(X) = =fa(2) da =/ Using the test from p-integrals from calculus, we know that this improper integral will only converge to a finite number when a > + . In this case we can find E(X) = Note: you can type o 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 00 E(X2 ) = -dr. Like before we know this will only converge to a finite number when a - 1 > + or in other words, when a > 4 Calculating we get E(X2) = From this we can calculate that Var(X) = (a-2)(0-1)2Let's calculate the mean and variance of the Gamma distribution. the probability density function of the Gamma distribution is defined to be fn(x) = mphe d for 1 2 0 0 otherwise. In general, the mean of a continuous random variable is defined to be E(X) = Ifn(I) di. So, in the case of the gamma distribution this is E(X) = 1antle Ida. Using a well know result (see note below) we can easily calculate this to be E(X) = The variance can be calculate by Var (X) = E ((X - E(X)) 2) , 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 E(X2) = DO Use the same result as before (see note below) we can easily calculate this to be E(X?) = From this we calculate that Var(X) = E(X2) - E(X)? = Note: Recall that important formula o me "de = m! for m = 0, 1, 2, 3