91影视

Variance is the expected squared variation of a random variable from its population mean or sample mean in probability theory and statistics. Variance is a measure of dispersion, or how far a set of numbers deviates from its average value.

Determine the integral of the joint pdf over the given region when \(0 \le x \le 20\)(which is then bounded by the two straight lines):

\(\begin{aligned}\int_0^{20} {\int_{20 - x}^{30 - x} f } (x,y)dydx &= \int_0^{20} {\int_{20 - x}^{30 - x} k } xydydx\\ = \left. {\int_0^{20} {\left( {\frac{{kx{y^2}}}{2}} \right)} } \right|_{20 - x}^{30 - x}dx &= \int_0^{20} {\left( {\frac{{kx{{(30 - x)}^2}}}{2} - \frac{{kx{{(20 - x)}^2}}}{2}} \right)} dx\\ = \int_0^{20} 2 50kx - 10k{x^2}dx &= \left. {\left( {125k{x^2} - \frac{{10}}{3}k{x^3}} \right)} \right|_0^{20}\\ &= 125k{(20)^2} - \frac{{10}}{3}k{(20)^3} = \frac{{70000}}{3}k\end{aligned}\)

Determine the integral of the joint pdf over the given region when \(20 \le x \le 30\)(which is then bounded by the top line and the x-axis):

\(\begin{aligned}\int_{20}^{30} {\int_0^{30 - x} f } (x,y)dydx &= \int_{20}^{30} {\int_0^{30 - x} k } xydydx\\ &= \left. {\int_{20}^{30} {\left( {\frac{{kx{y^2}}}{2}} \right)} } \right|_0^{30 - x}dx = \int_0^{20} {\frac{{kx{{(30 - x)}^2}}}{2}} dx\\ &= \int_{20}^{30} 4 50kx - 30k{x^2} + \frac{1}{2}k{x^3}dx = \left. {\left( {225k{x^2} - 10k{x^3} + \frac{1}{8}k{x^4}} \right)} \right|_{20}^{30}\\ &= 3750k\end{aligned}\)

The total integral of the joint pdf over the indicated region is then the sum of these two areas:

\(\frac{{70000}}{3}k + 3750k = \frac{{81250}}{3}k\)

The joint pdf is only a valid pdf if the total integral is equal to \(1\):

\(\frac{{81250}}{3}k = 1\)

Divide each side by \(\frac{{81250}}{3}\):

\(k = \frac{3}{{81250}}\)

(b) The marginal pdf of X is the integral of the joint pdf over all possible values of Y:

First let \(0 \le x \le 20\)(region is then bounded by the two straight lines):

\(\begin{aligned}{f_X}(x) &= \int_{20 - x}^{30 - x} f (x,y)dy &= \int_{20 - x}^{30 - x} k xydy\\ &= \left. {\left( {\frac{{kx{y^2}}}{2}} \right)} \right|_{20 - x}^{30 - x} = \left( {\frac{{kx{{(30 - x)}^2}}}{2} - \frac{{kx{{(20 - x)}^2}}}{2}} \right)\\ &= 250kx - 10k{x^2}\end{aligned}\)

Next let \(20 < x \le 30\)(region is then bounded by the top line and the $x$-axis):

\(\begin{array}{c}{f_X}(x) = \int_0^{30 - x} f (x,y)dy = \int_0^{30 - x} k = \left. {\left( {\frac{{kx{y^2}}}{2}} \right)} \right|_0^{30 - x}\\ = \frac{{kx{{(30 - x)}^2}}}{2}\\ = 450kx - 30k{x^2} + \frac{1}{2}k{x^3}\end{array}\)

Combining these results, we then obtain:

\(\begin{array}{l}{f_X}(x) = \left\{ {\begin{array}{*{20}{c}}{250kx - 10k{x^2}}&{0 \le x \le 20}\\{450kx - 30k{x^2} + \frac{1}{2}k{x^3}}&{20 < x \le 30}\end{array}} \right.\\\end{array}\)

The marginal distribution of Y is the same as the marginal distribution of X, because the joint pdf does not change when you interchange x and y (neither does the region).

\({f_Y}(y) = \left\{ {\begin{array}{*{20}{c}}{250ky - 10k{y^2}}&{0 \le y \le 20}\\{450ky - 30k{y^2} + \frac{1}{2}k{y^3}}&{20 < y \le 30}\end{array}} \right.\)

X and Y are not independent, because if you multiply the marginal distributions, then you do not obtain the joint pdf.

(c) Determine the integral of the joint pdf over the given region when \(0 \le x \le 20\)(which is then bounded by two straight lines):

\(\begin{array}{c}P(X + Y \le 25\mid 0 \le X \le 20) = \int_0^{20} {\int_{20 - x}^{25 - x} f } (x,y)dydx\\ = \int_0^{20} {\int_{20 - x}^{25 - x} k } xydydx = \left. {\int_0^{20} {\left( {\frac{{kx{y^2}}}{2}} \right)} } \right|_{20 - x}^{25 - x}dx\\ = \int_0^{20} {\left( {\frac{{kx{{(25 - x)}^2}}}{2} - \frac{{kx{{(20 - x)}^2}}}{2}} \right)} dx\\ = \int_0^{20} {\frac{{225}}{2}} kx - 5k{x^2}dx = \left. {\left( {\frac{{225}}{4}k{x^2} - \frac{5}{3}k{x^3}} \right)} \right|_0^{20}\\ = 125k{(20)^2} - \frac{{10}}{3}k{(20)^3} = \frac{{27500}}{3}k\end{array}\)

Determine the integral of the joint pdf over the given region when \(20 \le x \le 30\)(which is then bounded by the top line and the x-axis):

\(\)\(\begin{array}{c}P(X + Y \le 25\mid 20 < X \le 30) = \int_{20}^{25} {\int_0^{25 - x} f } (x,y)dydx\\ = \int_{20}^{25} {\int_0^{25 - x} k } xydydx = \left. {\int_{20}^{25} {\left( {\frac{{kx{y^2}}}{2}} \right)} } \right|_0^{25 - x}dx\\ = \int_0^{25} {\frac{{kx{{(25 - x)}^2}}}{2}} dx = \int_{20}^{25} {\frac{{625}}{2}} kx - 25k{x^2} + \frac{1}{2}k{x^3}dx\\ = \left. {\left( {\frac{{625}}{4}k{x^2} - \frac{{25}}{3}k{x^3} + \frac{1}{8}k{x^4}} \right)} \right|_{20}^{25} = \frac{{10625}}{{24}}k\end{array}\)

The total probability is then the sum of the two probabilities:

\(P(X + Y \le 25) = \frac{{27500}}{3}k + \frac{{10625}}{{24}}k = \frac{{76875}}{8}k = \frac{{76875}}{8}\frac{3}{{81250}} = \frac{{369}}{{1040}} \approx 0.3548 = 35.48\% \)

(d) The total amount is X+Y.

The expected value of X+Y is then integral of the product of all possible values \((x + y)\)and the joint pdf.

Determine the integral of the joint pdf over the given region when \(0 \le x \le 20\)(which is then bounded by the two straight lines):

\(\begin{array}{c}\int_0^{20} {\int_{20 - x}^{30 - x} {(x + y)} } f(x,y)dydx = \int_0^{20} {\int_{20 - x}^{30 - x} k } {x^2}y + kx{y^2}dydx\\ = \left. {\int_0^{20} {\left( {\frac{{k{x^2}{y^2}}}{2} + \frac{{kx{y^3}}}{3}} \right)} } \right|_{20 - x}^{30 - x}dx\\ = \int_0^{20} {\left( {\frac{{k{x^2}{{(30 - x)}^2}}}{2} + \frac{{kx{{(30 - x)}^3}}}{3} - \frac{{k{x^2}{{(20 - x)}^2}}}{2} - \frac{{kx{{(20 - x)}^3}}}{3}} \right)} dx\\ = \int_0^{20} {\frac{{19000}}{3}} kx - 250k{x^2}dx = \left. {\left( {\frac{{19000}}{6}k{x^2} - \frac{{250}}{3}k{x^3}} \right)} \right|_0^{20}\\ = 125k{(20)^2} - \frac{{10}}{3}k{(20)^3} = 600000k\end{array}\)

Determine the integral of the joint pdf over the given region when \(20 \le x \le 30\)(which is then bounded by the top line and the $x$-axis):

\(\begin{array}{c}\int_{20}^{30} {\int_0^{30 - x} {(x + y)} } f(x,y)dydx = \int_{20}^{30} {\int_0^{30 - x} k } {x^2}y + kx{y^2}dydx\\ = \left. {\int_{20}^{30} {\left( {\frac{{k{x^2}{y^2}}}{2} + \frac{{kx{y^3}}}{3}} \right)} } \right|_0^{30 - x}dx = \int_0^{20} {\frac{{k{x^2}{{(30 - x)}^2}}}{2}} + \frac{{kx{{(30 - x)}^3}}}{3}dx\\ = \int_{20}^{30} 9 000kx - 450k{x^2} + \frac{1}{6}k{x^4}dx = \left. {\left( {4500k{x^2} - 150k{x^3} + \frac{1}{{30}}k{x^5}} \right)} \right|_{20}^{30}\\ = \frac{{310000}}{3}k\end{array}\)

The expected total amount is then the sum of these two expected values:

\(E(X + Y) = 600000k + \frac{{310000}}{3}k = \frac{{2110000}}{3}k = \frac{{2110000}}{3}\frac{3}{{81250}} = \frac{{1688}}{{65}} \approx 25.9692\)

(e) Determine the following expected values in a similar manner as in part (a):

MEAN

\(\begin{array}{c}{\mu _X} = E(X) = \int_0^{30} x {f_X}(x)dx\\ = \int_0^{20} 2 50k{x^2} - 10k{x^3}dx + \int_{20}^{30} 4 50k{x^2} - 30k{x^3}\\ = \frac{{800000}}{3}k + 85000k = \frac{{1055000}}{3}k\\{\mu _Y} = E(Y) = E(X) = \frac{{1055000}}{3}k\end{array}\)

EXPECTED VALUE OF XY

\(\begin{array}{c}E(XY) = \int_0^{20} {\int_{20 - x}^{30 - x} x } yf(x,y)dydx + \int_{20}^{30} {\int_0^{30 - x} x } yf(x,y)dydx\\ = \int_0^{20} {\int_{20 - x}^{30 - x} k } {x^2}{y^2}dydx + \int_{20}^{30} {\int_0^{30 - x} k } {x^2}{y^2}dydx\\ = \frac{{29600000k}}{9} + \frac{{3650000k}}{9}\\ = \frac{{33250000k}}{9}\end{array}\)

VARIANCE

\(\begin{array}{c}\sigma _X^2 = E\left( {{{\left( {X - {\mu _X}} \right)}^2}} \right) = \int_0^{30} {{{\left( {x - \frac{{1055000}}{3}k} \right)}^2}} {f_X}(x)dx\\ = \int_0^{20} {{{\left( {x - \frac{{1055000}}{3}k} \right)}^2}} \left( {250kx - 10k{x^2}} \right)dx + \int_{20}^{30} {{{\left( {x - \frac{{1055000}}{3}k} \right)}^2}} \left( {450kx - 30k{x^2} + \frac{1}{2}k{x^3}} \right)dx\\ = \frac{{400000}}{{27}}k\left( {243 - 12660000k + 194779375000{k^2}} \right) + \frac{{25000}}{3}k\left( {233 - 7174000k + 55651250000{k^2}} \right)\\ = \frac{{16625000k}}{3} - \frac{{2226050000000{k^2}}}{9} + \frac{{90433281250000000{k^3}}}{{27}}\\\sigma _Y^2 = \sigma _X^2 = \frac{{16625000k}}{3} - \frac{{2226050000000{k^2}}}{9} + \frac{{90433281250000000{k^3}}}{{27}}\end{array}\)

The covariance is the expected value of X Y decreased by the product of the two means:

\(\begin{array}{c}{\mathop{\rm Cov}\nolimits} (X,Y) = E(XY) - {\mu _X}{\mu _Y} = \frac{{33250000k}}{9} - \frac{{1055000}}{3}\frac{{1055000}}{3}{k^2}\\ = \frac{{33250000}}{9}\frac{3}{{81250}} - \frac{{1055000}}{3}\frac{{1055000}}{3}\frac{{{3^2}}}{{{{81250}^2}}}\\ = - \frac{{408008}}{{12675}} \approx - 32.1900\end{array}\)

The correlation is the covariance divided by the product of the standard deviations. The standard deviations are the square root of the variances:\(\begin{array}{c}{\mathop{\rm Corr}\nolimits} (X,Y) = \frac{{{\mathop{\rm Cov}\nolimits} (X,Y)}}{{{\sigma _X}{\sigma _Y}}} = \frac{{ - \frac{{408008}}{{12675}}}}{{\frac{{16625000k}}{3} - \frac{{2226050000000{k^2}}}{9} + \frac{{90433281250000000{k^3}}}{{27}}}}\\ = \frac{{\frac{{16625000}}{3}\frac{3}{{81250}} - \frac{{2226050000000}}{9}\frac{{{3^2}}}{{{{81250}^2}}} + \frac{{90433281250000000}}{{27}}\frac{{{3^3}}}{{{{81250}^3}}}}}{9}\\ = - \frac{{102002}}{{114123}} \approx - 0.8938\end{array}\)

Note: The results of part (e) deviate from the result in the back of the book.

(f) The variance is the expected value of the deviation of the mean. The mean was determined in part (d). This expected value can be determined similarly:

\(\begin{array}{c}{\mathop{\rm Var}\nolimits} (X + Y) = \int_0^{20} {\int_{20 - x}^{30 - x} {{{(x + y - 25.9692)}^2}} } f(x,y)dydx + \int_{20}^{30} {\int_0^{30 - x} {{{(x + y - 25.9692)}^2}} } f(x,y)dydx\\ = \int_0^{20} {\int_{20 - x}^{30 - x} {{{(x + y - 25.9692)}^2}} } kxydydx + \int_{20}^{30} {\int_0^{30 - x} {{{(x + y - 25.9692)}^2}} } kxydydx \approx 184056k + 23141k\\ = 207197k = 207197\frac{3}{{81250}} = \frac{{621591}}{{81250}} \approx 7.6504\end{array}\)