Answer to Question #339373 in Statistics and Probability for ken

Question

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 ]

Answer 1

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

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

Thus, it is not a valid joint probability function.

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

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

Consider separately each probability in the last formula: $P(\frac13<X<\frac23,Y>\frac34)=\int_{\frac13}^{\frac23}\int_{\frac34}^{1}\frac25(2x+3y)dydx=\frac25\int_{\frac13}^{\frac23}(2xy+\frac32y^2)_{\frac34}^{1}dx=\frac25\int_{\frac13}^{\frac23}(2x+\frac32-\frac32x-\frac{27}{32})dx=\frac25\int_{\frac13}^{\frac23}(\frac12x+\frac{21}{32})dx=\frac25(\frac14x^2+\frac{21}{32}x)_{\frac13}^{\frac23}=\frac{29}{240}$

$P(Y>\frac34)=\frac25\int_0^1\int_{\frac34}^1(2x+3y)dydx=\frac25\int_0^1(2xy+\frac32y^2)|_{\frac34}^1dx=\frac25\int_0^1(2x+\frac32-\frac32x-\frac{27}{32})dx=\frac25\int_0^1(\frac12x+\frac{21}{32})dx=\frac25\left(\frac{1}{4}x^2+\frac{21}{32}x\right)|_0^1=\frac25\left(\frac14+\frac{21}{32}\right)=\frac{29}{80}$

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

Answer: $P(\frac13<X<\frac23,Y>\frac34)=\frac13$