Question #339373

1) The jdf of the random variables X, Y is given as

f(x, y) = { 2 /5 (2x + 3y), 0 < x < 1, 0 < y < 1

0, elsewhere

find Pr [ 1 /3 < X < 2 /3 |y > 3 /4 ]


1
Expert's answer
2022-05-11T12:06:22-0400

1). Since it is not clear how the joint probability function looks like, consider two cases:

a). f(x,y)=25(2x+3y)f(x,y)=\frac{2}{5(2x+3y)}. The following condition must be satisfied: R2f(x,y)dydx=1\iint_{\mathbb{R}^2}f(x,y)dydx=1. We receive: 25010112x+3ydydx=250113ln(2x+3y)01dx=21501(ln(2x+3)ln(2x))dy.\frac25\int_0^1\int_0^1\frac{1}{2x+3y}dydx=\frac25\int_0^1\frac13\ln(2x+3y)|_0^1dx=\frac{2}{15}\int_0^1(\ln(2x+3)-ln(2x))dy. Consider separately integral: ln(t)dt=tln(t)dt=tln(t)t+C\int\ln(t)dt=t\,\ln(t)-\int dt=t\ln(t)-t+C. Using the latter, we get: 25010112x+3ydydx=21501(ln(2x+3)ln(2x))dx=215(12(2x+3)ln(2x+3)32xln(2x))010.224\frac25\int_0^1\int_0^1\frac{1}{2x+3y}dydx=\frac{2}{15}\int_0^1(\ln(2x+3)-ln(2x))dx=\frac{2}{15}\left(\frac12(2x+3)\ln(2x+3)-\frac32-x\,\ln(2x)\right)|_0^1\approx0.224

Thus, it is not a valid joint probability function.

b). Consider the function: f(x,y)=25(2x+3y)f(x,y)=\frac25(2x+3y). We will get: 250101(2x+3y)dydx=2501(2xy+32y2)01dx=2501(2x+32)dx=25(x2+32x)01=1\frac25\int_0^1\int_0^1(2x+3y)dydx=\frac25\int_0^1(2xy+\frac32y^2)|_0^1dx=\frac25\int_0^1(2x+\frac32)dx=\frac25(x^2+\frac32x)|_0^1=1. Thus, it is a valid joint probability function.

Remind the formula: P(AB)=P(AB)P(B)P(A|B)=\frac{P(A\cap B)}{P(B)}. From the latter we get: P(13<X<23Y>34)=P(13<X<23,Y>34)P(Y>34)P(\frac13<X<\frac23|Y>\frac34)=\frac{P(\frac13<X<\frac23,Y>\frac34)}{P(Y>\frac34)}.

Consider separately each probability in the last formula: P(13<X<23,Y>34)=132334125(2x+3y)dydx=251323(2xy+32y2)341dx=251323(2x+3232x2732)dx=251323(12x+2132)dx=25(14x2+2132x)1323=29240P(\frac13<X<\frac23,Y>\frac34)=\int_{\frac13}^{\frac23}\int_{\frac34}^{1}\frac25(2x+3y)dydx=\frac25\int_{\frac13}^{\frac23}(2xy+\frac32y^2)_{\frac34}^{1}dx=\frac25\int_{\frac13}^{\frac23}(2x+\frac32-\frac32x-\frac{27}{32})dx=\frac25\int_{\frac13}^{\frac23}(\frac12x+\frac{21}{32})dx=\frac25(\frac14x^2+\frac{21}{32}x)_{\frac13}^{\frac23}=\frac{29}{240}

P(Y>34)=2501341(2x+3y)dydx=2501(2xy+32y2)341dx=2501(2x+3232x2732)dx=2501(12x+2132)dx=25(14x2+2132x)01=25(14+2132)=2980P(Y>\frac34)=\frac25\int_0^1\int_{\frac34}^1(2x+3y)dydx=\frac25\int_0^1(2xy+\frac32y^2)|_{\frac34}^1dx=\frac25\int_0^1(2x+\frac32-\frac32x-\frac{27}{32})dx=\frac25\int_0^1(\frac12x+\frac{21}{32})dx=\frac25\left(\frac{1}{4}x^2+\frac{21}{32}x\right)|_0^1=\frac25\left(\frac14+\frac{21}{32}\right)=\frac{29}{80}

Thus, the probability is the following: P(13<X<23Y>34)=292408029=13.P(\frac13<X<\frac23|Y>\frac34)=\frac{29}{240}\cdot\frac{80}{29}=\frac13.

Answer: P(13<X<23,Y>34)=13P(\frac13<X<\frac23,Y>\frac34)=\frac13


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