Consider randomly selecting a point (X1, X2, X3) in the unit cube {(x1, x2, x3) : 0<x1<1. f(x1,

Consider randomly selecting a point (X1, X2, X3) in the unit...

Consider randomly selecting a point $\left(X_{1}, X_{2}, X_{3}\right)$ in the unit cube $\left\{\left(x_{1}, x_{2}, x_{3}\right) : 0<x_{1}<1\right.$ $f\left(x_{1}, x_{2}, x_{3}\right)=\left\{\begin{array}{cc}{8 x_{1} x_{2} x_{3}} & {0<x_{1}<1, \quad 0<x_{2}<1, \quad 0<x_{3}<1} \\ {0} & {\text { otherwise }}\end{array}\right.$ (so the three variables are independent). Then form a rectangular solid whose vertices are $(0,0,$ 0), $\left(x_{1}, 0,0\right),\left(0, X_{2}, 0\right),\left(X_{1}, X_{2}, 0\right),\left(0,0, X_{3}\right),\left(X_{1}, 0, X_{3}\right),\left(0, X_{2}, X_{3}\right),$ and $\left(X_{1}, X_{2}, X_{3}\right) .$ The $Y_{2}=X_{1} X_{2} \cdot ]$

Consider randomly selecting a point $\left(X_{1}, X_{2}, X_{3}\right)$ in the unit cube $\left\{\left(x_{1}, x_{2}, x_{3}\right) : 0<x_{1}<1\right.$
$f\left(x_{1}, x_{2}, x_{3}\right)=\left\{\begin{array}{cc}{8 x_{1} x_{2} x_{3}} & {0<x_{1}<1, \quad 0<x_{2}<1, \quad 0<x_{3}<1} \\ {0} & {\text { otherwise }}\end{array}\right.$
(so the three variables are independent). Then form a rectangular solid whose vertices are $(0,0,$
0), $\left(x_{1}, 0,0\right),\left(0, X_{2}, 0\right),\left(X_{1}, X_{2}, 0\right),\left(0,0, X_{3}\right),\left(X_{1}, 0, X_{3}\right),\left(0, X_{2}, X_{3}\right),$ and $\left(X_{1}, X_{2}, X_{3}\right) .$ The
$Y_{2}=X_{1} X_{2} \cdot ]$

Key Concepts

Probability Density Function

A probability density function (pdf) describes how the probability is distributed over the values of a continuous random variable. It provides a function whose integral over an interval gives the probability that the variable falls within that interval, and it must satisfy the normalization condition that the total area under the curve equals one.

Independence of Random Variables

Independence means that the occurrence of one event or the value of one random variable does not affect the probability distribution of another. When applying this to a joint pdf, it implies that the function factors into the product of the marginal pdfs of the individual variables, allowing for separate analysis of each component.

Marginal Distribution

The marginal distribution of a random variable is obtained by integrating the joint probability density function over the other variables. This gives the probability distribution of that variable in isolation, without the influence of the others, and is essential when dealing with multi-dimensional probability problems.

Transformation of Random Variables

Transforming random variables involves finding the distribution of a new variable that is defined as a function of one or more other random variables. This often requires techniques such as the change-of-variable method and the use of the Jacobian determinant when the transformation is non-linear.

Multiple Integration

Multiple integration is used to compute probabilities or expectations in multi-dimensional spaces. By integrating a joint probability density function over a defined region, one can determine the probability that the random variables lie within that region, which is particularly useful when working with problems involving geometric shapes or multiple random variables.

Transcript

00:03 We randomly select a points x1, x2, x3, and a unit cube, according to a given print probability density function in terms of the three coordinates x1x2 and x3.

00:27 Then we form a rectangular solid with the vertices 0 to the origin, x1 .00, x1, x20, x1x20, x1x2, 00x3, x10x3, x10x3, x10x3, x10x3, 0 x2 x3 and x1 x2 x3 and of course the volume of this cube which we're calling a r of 3 is going to be x1 times x2 times x3 product of the three different size and we're asked to find the probably density function of the volume so it's suggesting this problem that we we should take y1 equal to x1 and y2 should be equal to x1 times x2.

01:36 And then we have, as we said before, y3 is x1, x2, x3.

01:46 Then you want to solve for the xs in terms of the y's.

01:51 We have it again, x1 is equal to y1.

01:57 We have the x2, so this is y2 over y1, sorry, and then x3, this is y3 over x1, which is y2, so this is y3 over y2.

02:27 Now we have the jacobian of the transformation is going to be the determinants of matrix, with entries, where we have partial of x1 with respect to y1, is 1.

02:45 All the other partials are 0.

02:49 The partial with x2 with respect to y2 is 1 over y1.

02:56 The partial of x2 is 2, sorry, except to y1, i'm sorry, this is negative y2 over y1 squared.

03:07 And then with respect to y2 is 1 over y1.

03:11 Expect the y3 is 0.

03:13 The partial of x2 with respect to y1 is 0 with respect to y2 is negative y3 over y2 squared.

03:24 Expect the y3 is 1 over y3.

03:29 Now we see this is upper triangular, so it follows that the determinants of this matrix simply go into the product of the diagonals, which is 1 over y1 times this should be one of the y2 here.

03:53 So one over y1, y2 here.

03:55 So 1 over y1, y2.

03:56 So the joint probability density function at the y's, f of y1, f of y2, f of y 3, well we were given the probability density function of x is 8 x1 x to x3.

04:22 So substituting we get 8 x times called x1 is y1, x2 is y2 if y1.

04:33 So it's 2, y2 or y1.

04:35 So it's y2 or y1.

04:38 Times x3 which is y3 over y2...

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