This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x=1, the value of the joint PDF is 1/2.
(figure belongs to "the science of uncertainty)
1. Are X and Y independent? NO
2. Find fX(x). Express your answers in terms of x using standard notation .
If 0<x<1,
fX(x)= x/2
If 1<x<2,
fX(x)= -3*x/2+3
Find fY|X(y∣0.5).
If 0<y<1/2,
fY|X(y∣0.5)= 2
3. Find fX|Y(x∣0.5).
If 1/2<x<1,
fX|Y(x∣0.5)= 0.5
If 1<x<3/2,
fX|Y(x∣0.5)= 1.5
Let R=XY and let A be the event {X<0.5}. Evaluate E[R∣A].
E[R∣A]= 0.0625
Answered in full...
Sophia is vacationing in Monte Carlo. On any given night, she takes X dollars to the casino and returns with Y dollars. The random variable Xhas the PDF shown in the figure. Conditional on X=x, the continuous random variable Y is uniformly distributed between zero and 3x.
This figure below describes the joint PDF of the random variables X and Y. These random variables take values in [0,2] and [0,1], respectively. At x=1, the value of the joint PDF is 1/2.
1. Are X and Y independent?
No
2. Find fX(x). Express your answers in terms of x, using the standard notation.
If 0<x≤1:
fX(x)= 1/2⋅x
If 1<x<2:
fX(x)= −3⋅x/2+3
If x<0 or x≥2:
fX(x)= 0
3. Find fY∣X(y∣0.5).
If 0<y<1/2:
fY|X(y∣0.5)= 2
If y<0 or y>1/2:
fY|X(y∣0.5)= 0
4. Find fX∣Y(x∣0.5).
If 1/2<x<1:
fX|Y(x∣0.5)=
If 1<x<3/2:
fX|Y(x∣0.5)= 1.5
If x<1/2 or x>3/2:
fX|Y(x∣0.5)= 0
5. Let R=XY and let A be the event that {X<0.5}. Find E[R|A].
E[R∣A]= 0.0625
4. Find fX∣Y(x∣0.5).
If 1/2<x<1:
fX|Y(x∣0.5)=0.5
4. Find fX(x).
If 1<x<2:
fX(x) = (-3)*x/2+3
E[R∣A]= 0.0625
Well, that's not a very high expectation, but at least it's a positive one! Though, maybe it's time for X to step up its game and strive for bigger values. Keep your chin up, X!
To determine if X and Y are independent, the joint probability density function (PDF) is analyzed. In this case, it is given that at x=1, the value of the joint PDF is 1/2. This value doesn't depend on the value of y and is not equal to the product of the marginal PDFs of X and Y. Therefore, X and Y are not independent.
To find the marginal PDF of X, we need to integrate the joint PDF of X and Y with respect to y. In this case, the joint PDF is given in two different intervals of x.
For 0<x<1, the marginal PDF of X (fX(x)) is found by integrating the joint PDF with respect to y over the range [0,1]. This gives us fX(x) = x/2.
For 1<x<2, the marginal PDF of X is found by integrating the joint PDF with respect to y over the range [0,1]. This gives us fX(x) = -3*x/2+3.
To find the conditional PDF of Y given X=0.5 (fY|X(y|0.5)), we consider X=0.5 as a condition and find the conditional PDF of Y. In this case, it is given that the conditional PDF is 2 for 0<y<1/2.
To find the conditional PDF of X given Y=0.5 (fX|Y(x|0.5)), we consider Y=0.5 as a condition and find the conditional PDF of X. In this case, it is given that the conditional PDF is 0.5 for 1/2<x<1 and 1.5 for 1<x<3/2.
To evaluate E[R|A], where R=XY and A is the event that X<0.5, we need to find the conditional expectation of R given A. By using the definition of conditional expectation, we calculate the expected value of R given that X<0.5. In this case, the value of E[R|A] is 0.0625.
Therefore, all the answers are provided according to the calculated values.