The probability density function of a Pareto distribution...

The probability density function of a Pareto distribution is: f_X(x) = { ? / x^(?+1) , x ? 1 0 , x < 1 where ? > 0. The mean of a Pareto distribution is E(X) = ? / (? - 1) , ? > 1, and the variance is Var(X) = ? / ((? - 1)^2 * (? - 2)) , ? > 2. (a) Compute E[X^4]. [4 marks] (b) Calculate the cumulative distribution function when ? = 4. [4 marks] (c) Using the result in part (b), calculate Pr(X > 2) and Pr(log X ? 3). [4 marks] (d) Let Y_i be a Pareto distribution with ? = i + 2, i = 1, 2, 3. Y_1 and Y_2 are independent and Cov(Y_2, Y_3) = 1/8. (i) Let Y = 3Y_1 + 2Y_2. Compute the mean and variance of Y. [4 marks] (ii) Let Z = 5Y_2 + 4Y_3. Compute the mean and variance of Z. [4 marks]

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00:01 So, the pdf of the parieto distribution is given by alpha by x power alpha plus 1, x greater than it equal to 1, and 0 when x is less than 1.

00:18 This is the probability density function of the parity 2 distribution.

00:22 Now let's find expectation of x to the power of 4.

00:25 If you can see the support of brand available x is greater than it equal to 1, we have the non -zero part.

00:31 But for less than one there is a zero part so we can start directly number one the low limit is one and the upper limit is infinity x power four times the pd alpha by x power alpha plus one d x so alpha you can take it out one to infinity x power if you subtract the powers it is t minus alpha d alpha so that means it is alpha into what is the integration of x power three minus alpha it is x power four minus alpha divided by four minus alpha assuming alpha or equal to 4 then substitute the limits 1 to infinity now when you substitute infinity it will be 0 provided alpha is greater than 4 so alpha should be greater than 4 then only the upper limit tends to 0 so 0 and when you substitute the low limit it is negative alpha by 4 minus alpha or alpha by alpha minus 4.

01:35 So, expectation of x to the power of 4 is basically alpha by alpha minus 4 alpha greater than 4.

01:43 This is the answer.

01:45 For the first question, when alpha is 4, what is the period of this random variable when alpha is 4? so when alpha is 4, we have 4 by x to the power of 5 x greater than are equal to 1 and 0 when x is less than 1.

02:01 When this is the probability density function what is the cdf cumulative distribution function given by the cumulative distribution function is basically negative infinity to x the pdf or you can take du fx of u du this is the definition of the cumulative distribution function cdf now we have two cases when x is when x is less than 1 the pdf is 0 so let's take those cases so case 1 when x is less than 1 cumulative distribution function will become negative infinity to x 0 d u and that will be 0 case 2 when x is greater than or equal to 1 then the cdf will become negative infinity to 1 0 d u because in this interval the pdf is taking 0 plus 1 to x now the pdf takes 4 by x to the power of 5 or u to the power of 5 4 by u to the power of 5 d u all right so let's do the integration so first integration is 0 the second integration is 4x par negative 5 what is integration of 4x par negative 5 is 4x par negative 4 by negative from 1 to x so substitute in the limits it will be 1 minus 1 by x to the power of 4 so the cdf of the random variable when alpha is 4 is given by 0 when x is less than 1 and 1 minus 1 by x to the power of 4 4 when x is greater than or equal to 1.

03:37 So this is the cdf.

03:39 Basically, it is probability that x is less than equal to x.

03:42 So this is the cdf of the random variable parity distribution.

03:46 Now, let's find probably that x is greater than 2.

03:48 So the probability that x is greater than 2 is basically 1 minus probability that x is less than or equal to 2.

03:55 That is 1 minus the cdf evaluated at 2.

03:58 That is 1 minus.

03:59 So when will be cdf evaluated at 2? we need to take the second piece.

04:03 So it is 1 minus 1 by 2 to the power of 4.

04:05 So, answer is 1 by 16.

04:10 How about probably that log x is less than or equal to 3? probably that log x is less than equal to this implies probably that x is less than or equal to, i'm assuming log s to the base e.

04:22 I'm not sure because it's not mentioned.

04:24 I'm assuming it is base e.

04:25 So it is log x to the, i mean, x is less than equal to e cube.

04:31 So that is the cdf evaluated at e cube.

04:34 So answer is 1 minus 1 by e to the power of 12.

04:40 Now it's given that y1 follows, y1 follows parito distribution with alpha 3, y2 follows parito distribution with alpha 4, and y3 follows, follows parito distribution with alpha 5...

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