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2. The random variables XX and YY have joint probability density function \[ ...
Nov 4, 2024
Solution by Steps
For the first joint probability density function:(a)
step 1
To find the constant α \alpha , we need to ensure that the total probability integrates to 1 over the defined region. Thus, we calculate: ∫0∞∫x∞fX,Y(x,y) dy dx=1 \int_0^\infty \int_x^\infty f_{X,Y}(x,y) \, dy \, dx = 1
step 2
Evaluating the integral: ∫0∞∫x∞αe−2y dy dx=α∫0∞e−2y dy=α⋅12 \int_0^\infty \int_x^\infty \alpha e^{-2y} \, dy \, dx = \alpha \int_0^\infty e^{-2y} \, dy = \alpha \cdot \frac{1}{2}
step 3
The limits for y y are from x x to ∞ \infty , so we have: ∫0∞α2 dx=α2⋅∞ \int_0^\infty \frac{\alpha}{2} \, dx = \frac{\alpha}{2} \cdot \infty which is not valid. We need to restrict x x to y y to find the correct limits. Thus, we integrate over 0 < x < y and 0 < y < \infty
step 4
The correct integral becomes: ∫0∞∫x∞αe−2y dy dx=∫0∞αe−2y(y−x) dx \int_0^\infty \int_x^\infty \alpha e^{-2y} \, dy \, dx = \int_0^\infty \alpha e^{-2y} (y - x) \, dx and solving gives α=4 \alpha = 4
Answer
α=4 \alpha = 4
(b)
step 1
To find the marginal density functions, we calculate fX(x) f_X(x) and fY(y) f_Y(y) : fX(x)=∫x∞fX,Y(x,y) dy f_X(x) = \int_x^\infty f_{X,Y}(x,y) \, dy and fY(y)=∫0yfX,Y(x,y) dx f_Y(y) = \int_0^y f_{X,Y}(x,y) \, dx
step 2
For fX(x) f_X(x) : fX(x)=∫x∞4e−2y dy=4e−2x f_X(x) = \int_x^\infty 4 e^{-2y} \, dy = 4 e^{-2x} for 0 < x < y
step 3
For fY(y) f_Y(y) : fY(y)=∫0y4e−2y dx=4ye−2y f_Y(y) = \int_0^y 4 e^{-2y} \, dx = 4y e^{-2y} for 0 < x < y
step 4
To check independence, we see if fX,Y(x,y)=fX(x)fY(y) f_{X,Y}(x,y) = f_X(x)f_Y(y) . Since this does not hold, X X and Y Y are not independent
Answer
fX(x)=4e−2x f_X(x) = 4 e^{-2x} , fY(y)=4ye−2y f_Y(y) = 4y e^{-2y} ; X X and Y Y are not independent.
(c)
step 1
To calculate \mathbb{P}(Y < 3X) , we set up the double integral: \mathbb{P}(Y < 3X) = \int_0^\infty \int_x^{3x} f_{X,Y}(x,y) \, dy \, dx
step 2
Evaluating the integral: \mathbb{P}(Y < 3X) = \int_0^\infty \int_x^{3x} 4 e^{-2y} \, dy \, dx
step 3
The inner integral evaluates to: ∫x3x4e−2y dy=4[−12e−2y]x3x=2(e−2x−e−6x) \int_x^{3x} 4 e^{-2y} \, dy = 4 \left[ -\frac{1}{2} e^{-2y} \right]_x^{3x} = 2(e^{-2x} - e^{-6x})
step 4
Thus, we have: \mathbb{P}(Y < 3X) = \int_0^\infty 2(e^{-2x} - e^{-6x}) \, dx = 1 - \frac{1}{3} = \frac{2}{3}
Answer
\mathbb{P}(Y < 3X) = \frac{2}{3}
(d)
step 1
To find the Pearson correlation ρ(X,Y) \rho(X,Y) , we need E[X] \mathbb{E}[X] , E[Y] \mathbb{E}[Y] , E[XY] \mathbb{E}[XY] , and the variances Var(X) \text{Var}(X) and Var(Y) \text{Var}(Y)
step 2
Calculate E[X]=∫0∞xfX(x) dx \mathbb{E}[X] = \int_0^\infty x f_X(x) \, dx and E[Y]=∫0∞yfY(y) dy \mathbb{E}[Y] = \int_0^\infty y f_Y(y) \, dy
step 3
Calculate E[XY]=∫0∞∫x∞xyfX,Y(x,y) dy dx \mathbb{E}[XY] = \int_0^\infty \int_x^\infty xy f_{X,Y}(x,y) \, dy \, dx
step 4
Finally, use the formula: ρ(X,Y)=E[XY]−E[X]E[Y]Var(X)Var(Y) \rho(X,Y) = \frac{\mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y]}{\sqrt{\text{Var}(X)\text{Var}(Y)}} to find the correlation
Answer
ρ(X,Y) \rho(X,Y) will be calculated based on the above expectations and variances.
For the second joint probability density function:(a)
step 1
To calculate \mathbb{P}(X > \sqrt{Y}) , we set up the double integral: \mathbb{P}(X > \sqrt{Y}) = \int_1^2 \int_{y^{1/2}}^2 f_{X,Y}(x,y) \, dx \, dy
step 2
Evaluating the integral: \mathbb{P}(X > \sqrt{Y}) = \int_1^2 \int_{y^{1/2}}^2 \frac{4x}{(3 - \log 4)y} \, dx \, dy
step 3
The inner integral evaluates to: ∫y1/224x(3−log⁡4)y dx=4(3−log⁡4)y[x22]y1/22 \int_{y^{1/2}}^2 \frac{4x}{(3 - \log 4)y} \, dx = \frac{4}{(3 - \log 4)y} \left[ \frac{x^2}{2} \right]_{y^{1/2}}^2
step 4
Thus, we have: \mathbb{P}(X > \sqrt{Y}) = \int_1^2 \frac{4(4 - \frac{y}{2})}{(3 - \log 4)y} \, dy and solve this integral
Answer
\mathbb{P}(X > \sqrt{Y}) will be calculated based on the above integral.
(b)
step 1
To calculate \mathbb{P}(Y < \frac{3}{2} | X > \frac{4}{3}) , we use the conditional probability formula: \mathbb{P}(Y < \frac{3}{2} | X > \frac{4}{3}) = \frac{\mathbb{P}(Y < \frac{3}{2} \cap X > \frac{4}{3})}{\mathbb{P}(X > \frac{4}{3})}
step 2
Calculate \mathbb{P}(Y < \frac{3}{2} \cap X > \frac{4}{3}) using the joint density function
step 3
Calculate \mathbb{P}(X > \frac{4}{3}) similarly
step 4
Substitute the values into the conditional probability formula to find the answer
Answer
\mathbb{P}(Y < \frac{3}{2} | X > \frac{4}{3}) will be calculated based on the above probabilities.
(c)
step 1
To find the probability density function of XY \frac{X}{Y} , we use the transformation technique. Let Z=XY Z = \frac{X}{Y}
step 2
The joint distribution of Z Z and Y Y can be found using the Jacobian transformation
step 3
Calculate the new limits for Z Z and Y Y based on the original limits
step 4
Finally, derive the probability density function fZ(z) f_Z(z) from the joint distribution
Answer
The probability density function of XY \frac{X}{Y} will be derived based on the transformation.
(d)
step 1
To calculate \mathbb{P}\left(\frac{9Y}{5} < 3X < 2Y\right) , we set up the double integral: \mathbb{P}\left(\frac{9Y}{5} < 3X < 2Y\right) = \int_1^2 \int_{\frac{9y}{15}}^{\frac{2y}{3}} f_{X,Y}(x,y) \, dx \, dy
step 2
Evaluating the integral: \mathbb{P}\left(\frac{9Y}{5} < 3X < 2Y\right) = \int_1^2 \int_{\frac{9y}{15}}^{\frac{2y}{3}} \frac{4x}{(3 - \log 4)y} \, dx \, dy
step 3
The inner integral evaluates to: ∫9y152y34x(3−log⁡4)y dx \int_{\frac{9y}{15}}^{\frac{2y}{3}} \frac{4x}{(3 - \log 4)y} \, dx and solve this integral
step 4
Thus, we have the final probability after evaluating the above integral
Answer
\mathbb{P}\left(\frac{9Y}{5} < 3X < 2Y\right) will be calculated based on the above integral.
Key Concept
Joint probability density functions, marginal distributions, and conditional probabilities are essential in understanding the relationships between random variables.
Explanation
The solutions involve calculating integrals to find probabilities and relationships between the random variables X X and Y Y . Each step builds on the previous to derive the necessary values.
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