91影视

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

First, notice how the random variable \((X,Y)\)is evenly scattered over the disc. This suggests that we can compute the probability by calculating the area of a subset.

The following is true for every acceptable set \(A\):

\({\rm{P((X,Y)I A) = }}\mathop \int\!\!\!\int \nolimits_{\rm{A}} {\rm{nf(x,y)dxdy}}\)

Because \(A\)is a circle with radius\({\rm{R/2}}\)in our example, we obtain

\(\begin{aligned}{*{20}{c}}{{\rm{P((X,Y)I A) = }}\mathop \int\!\!\!\int \nolimits_{\rm{A}} {\rm{nf(x,y)dxdy = }}\mathop \int\!\!\!\int \nolimits_{\rm{A}} {\rm{n}}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{dxdy}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{ \times \pi }}{{\left( {\frac{{\rm{R}}}{{\rm{2}}}} \right)}^{\rm{2}}}}\\{{\rm{ = 0}}{\rm{.25}}{\rm{.}}}\end{aligned}\)

The integral is equal to the area of the circle.

First, represent the case when \(X\)departs from \({\rm{0}}\) by at most \({\rm{R/2}}\)as follows:

\({\rm{ - }}\frac{{\rm{R}}}{{\rm{2}}}{\rm{拢X拢}}\frac{{\rm{R}}}{{\rm{2}}}\)

because it has the ability to travel in both directions (negative and positive). We do the same for \(Y\). As a result, we require the likelihood of the given events intersecting

\(\begin{aligned}{*{20}{c}}{{\rm{P}}\left( {{\rm{ - }}\frac{{\rm{R}}}{{\rm{2}}}{\rm{拢X拢}}\frac{{\rm{R}}}{{\rm{2}}}{\rm{, - }}\frac{{\rm{R}}}{{\rm{2}}}{\rm{拢Y拢}}\frac{{\rm{R}}}{{\rm{2}}}} \right){\rm{ = }}\mathop \int\!\!\!\int \nolimits_{\rm{A}} {\rm{nf(x,y)dxdy}}}\\{{\rm{ = }}\mid {\rm{A\;is a square with side\;}}\frac{{\rm{R}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{R}}}{{\rm{2}}}{\rm{ = R}}\mid }\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{ \times }}{{\rm{R}}^{\rm{2}}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{\rm{\pi }}}}\end{aligned}\)

The probability can be calculated in the same way. The only difference is that the side length is longer. Therefore

\(\begin{aligned}{*{20}{c}}{{\rm{P}}\left( {{\rm{ - }}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}{\rm{拢X拢}}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}{\rm{, - }}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}{\rm{拢Y拢}}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}} \right){\rm{ = }}\mathop \int\!\!\!\int \nolimits_{\rm{A}} {\rm{nf(x,y)dxdy}}}\\{{\rm{ = }}\mid {\rm{A\;is a square with side\;}}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}{\rm{ + }}\frac{{\rm{R}}}{{\sqrt {\rm{2}} }}{\rm{ = }}\sqrt {\rm{2}} {\rm{R}}\mid }\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{ \times (}}\sqrt {\rm{2}} {\rm{R}}{{\rm{)}}^{\rm{2}}}}\\{{\rm{ = }}\frac{{\rm{2}}}{{\rm{\pi }}}{\rm{ \times }}}\end{aligned}\)

For the continuous random variable \({\rm{X}}\),the marginal probability density function is

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = `o }}_{{\rm{ - 楼}}}^{\rm{楼}}{\rm{nf(x,y)dy,\;for\; - 楼 < x < 楼}}\)

The integral's bounds are determined using

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = }}{{\rm{R}}^{\rm{2}}}\)

and we obtain

\({\rm{y = }}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{,}}\;\;\;{\rm{y = - }}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} \)

As a result, the following applies:

\(\begin{aligned}{*{20}{c}}{{{\rm{f}}_{\rm{X}}}{\rm{(x) = `o }}_{{\rm{ - 楼}}}^{\rm{楼}}{\rm{nf(x,y)dy = `o }}_{{\rm{ - }}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} }^{\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} }{\rm{n}}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{dy}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{ \times 2}}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}} {\rm{, - R拢x拢R}}}\end{aligned}\)

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = 0,}}\)

Similarly, $Y$'s marginal pdf is

\(\begin{aligned}{*{20}{c}}{{{\rm{f}}_{\rm{Y}}}{\rm{(y) = `o }}_{{\rm{ - 楼}}}^{\rm{楼}}{\rm{nf(x,y)dx = `o }}_{{\rm{ - }}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} }^{\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} }{\rm{n}}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{dx}}}\\{{\rm{ = }}\frac{{\rm{1}}}{{{\rm{\pi }}{{\rm{R}}^{\rm{2}}}}}{\rm{ \times 2}}\sqrt {{{\rm{R}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} {\rm{, - R拢y拢R,}}}\\{{{\rm{f}}_{\rm{X}}}{\rm{(y) = 0,\;otherwise}}{\rm{.\;}}}\end{aligned}\)

Two independent variables \(X and Y\)are independent if and only if

1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\), for every\({\rm{(x,y)}}\)and when \(X and Y\) discrete rv's, and

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\), for every \({\rm{(x,y)}}\), otherwise they are dependent

The joint pdf is obviously not equal to the product of the two marginal distributions , hence the random variables are dependent.