statistics questions

Problem (3.8.12)

Let ,

a.

In order to find the value of c, we have to integrate the double integration using the appropriate limits of integration and therefore we should set the result equal to the 0 and solve for c. at first we have to integrate with respect to y first, as it is given that, so it should be our limits for integration and for 0 and are our limits of integration when we integrate with respect to x.

So, 8c=1

c=1/8

Thus

b.

To find the marginal with respect to y, we integrate out x. The domain −x ≤ y ≤ x can be rewritten as y ≤ |x| ⇒ |y| ≤ |x| ⇒ |x| ≥ |y| ⇒ x ≥ |y| since x is non-negative. So we integrate from |y| to ∞

To find the marginal density of X, integrate out Y and choose limits of integration that result in the function in the terms of x. We have the domain is so we have to integrate from -x to x.

c.

The conditional distribution of (X,Y) given Z=z has the density

Problem (3.8.73)

Here are the independent and discrete random variables and each has the density function f.

In this case, P(=, =)=P(x1=x1)P(=)

Or equivalently we can say

Conversely, for all and x2 the joint probability function f(,x2) if it can be expressed as the product of the function of alone and a function of alone s(as they are the marginal probability function of and ), and are independent. We can say that cannot be so expressed then and are dependent.

So, if and are continuous random variables then we can say that they are tactually the independent variables if the and are independent events for all and .

So in this case we can write,

Or equivalently,

Where are the marginal distribution function of , . So, , and are independent random variables if for all ,and , their joint distribution function F(, can be expressed as the product of the function of and the function of as they are the marginal distribution of ,and , respectively. So, can’t be expressed then and are dependent.

For continuous independent random variables, it is true that the joint density function is the product of the function of individually and the other function of individually.

Problem (3.8.74)

a.

let, are the independent uniform random variables. We know that the joint density is the product of the individual functions.

Thus we can say that where 0<<1, 0<<1, 0<<1

So it can be said that the joint density is also uniform.

b.

The locations of three gas stations are independently and randomly placed along a mile of highway.

Let, the probability that two gas stations are less than 1/3 mile part is,

The answer of the above probability is 1/18

(1-1/18)=17/18=0.94

So we can say that the overall probability that no of two gas stations is actually less than 1/3 mile apart is 0.94.

4.

The sup Xnand lim supn→∞Xn are random variables because supremum(sup) is valid for all n, that means or we can write in this way like

( = sup {Xn( : n N}

So for this reason Sup Xn is random variable.

Similarly lim supXnis a random variable, for each . Thus

(= (.

That means for fixed, we get a sequence of real number i.e. {X1(, X2(…}. Thus lim supXnis also a random variable.

5.

a.

As we know that some of two independent chi-square variates are a chi-variate with a degree of freedom equal to the sum of the degrees of freedom of its additive components.

So if X1, X2…Xnis an independent random variable from a standard normal N(0, σ^2) 0r N(0,1) because ofσ^2=1 thenχ^2(1)

χ^2(n)

If we put the value of mean and variance i.e. N (0, 1) than we get the value of above equation such as

X12χ2(1)……………(1)

And also we know that if given variable are independent variable then

X12gamma( in given question

So X12gamma(…………………..(2)

Therefore from eqn (1) and (2), we conclude that

X12χ2(1) gamma(

b.

Density of Z= dX

c.

From Gaussian distribution theorem we know that for every and >0, Probability density function is

f(x) =

Where X is a continuous random variable with parameters and (XN (0,)), but in some special case and then =1. Therefore

f(x) = ……………………………(1)

When we try to plot the function with mean and the variance then the value of density becomes 1. i.e.

dx = dX = 1………………………………………..(2)

From both equation (1) and (2) we see that density function does not contain the σ term.