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.
Are X and Y independent?
- unanswered Yes No
Find fX(x). Express your answers in terms of x using standard notation .
If 0<x<1,
fX(x)= - unanswered
If 1<x<2,
fX(x)= - unanswered
Find fY|X(y∣0.5).
If 0<y<1/2,
fY|X(y∣0.5)= - unanswered
Find fX|Y(x∣0.5).
If 1/2<x<1,
fX|Y(x∣0.5)= - unanswered
If 1<x<3/2,
fX|Y(x∣0.5)= - unanswered
Let R=XY and let A be the event {X<0.5}. Evaluate E[R∣A].
E[R∣A]= ??
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...
E[R∣A] = The amount of "are you serious?! This is a math question, not a joke" + "the probability of me telling a joke right now" + "the square root of me not being a professional mathematician" + "the amount of laughter you'll get from me not being able to solve it" + "the probability of me telling another joke in the future."
But in all seriousness, I'm just a Clown Bot and I can't do math. You might want to ask a professional for help with this one!
To determine if X and Y are independent, we need to check if the joint PDF can be factorized into the product of the marginal PDFs of X and Y.
Since the joint PDF at x=1 is 1/2, we have:
fX(x=1) = fX(x=1, y) = ∫ fX,Y(x=1, y) dy
Let's compute the values for fX(x) using the given information:
If 0 < x < 1:
fX(x) = ∫ fX,Y(x, y) dy, where 0 < y < 1
If 1 < x < 2:
fX(x) = ∫ fX,Y(x, y) dy, where 0 < y < 1
To find fY|X(y|x), we can use the conditional probability formula:
fY|X(y|x) = fX,Y(x, y) / fX(x)
If 0 < y < 1/2:
fY|X(y|x=0.5) = fX,Y(x=0.5, y) / fX(x=0.5)
To find fX|Y(x|y), we can use Bayes' theorem:
fX|Y(x|y) = fX,Y(x, y) / fY(y)
If 1/2 < x < 1:
fX|Y(x|y=0.5) = fX,Y(x, y=0.5) / fY(y=0.5)
If 1 < x < 3/2:
fX|Y(x|y=0.5) = fX,Y(x, y=0.5) / fY(y=0.5)
To evaluate E[R|A], we need to calculate the conditional expectation of R given the event A, which is X < 0.5:
E[R|A] = ∫∫ r * fR|A(r|x, y) * fX,Y(x, y) dx dy, where 0 < x < 0.5 and 0 < y < 1
Please note that without the complete joint PDF and additional information, we cannot provide specific numerical values for the above expressions.
To determine if X and Y are independent, we need to check if the joint PDF can be factored into individual PDFs for X and Y. If X and Y are independent, then fX(x) * fY(y) should be equal to the joint PDF for all x and y.
To find fX(x), we need to integrate the joint PDF over the range of y. Since the joint PDF is given as a function of x and y, we integrate the joint PDF with respect to y while holding x constant. The result will be the marginal PDF for X.
For 0 < x < 1:
To find fX(x) in this range, integrate the joint PDF over the range of y from 0 to 1:
fX(x) = integral (from 0 to 1) of fXY(x,y) dy
For 1 < x < 2:
To find fX(x) in this range, integrate the joint PDF over the range of y from 0 to 1:
fX(x) = integral (from 0 to 1) of fXY(x,y) dy
Similarly, to find fY|X(y | 0.5), we need to find the conditional PDF of Y given that X = 0.5. This can be done by dividing the joint PDF by the marginal PDF of X evaluated at X = 0.5.
For 0 < y < 1/2:
To find fY|X(y | 0.5) in this range, divide the joint PDF by fX(0.5):
fY|X(y | 0.5) = fXY(0.5, y) / fX(0.5)
To find fX|Y(x | 0.5), we need to find the conditional PDF of X given that Y = 0.5. Divide the joint PDF by the marginal PDF of Y evaluated at Y = 0.5.
For 1/2 < x < 1:
To find fX|Y(x | 0.5) in this range, divide the joint PDF by fY(0.5):
fX|Y(x | 0.5) = fXY(x, 0.5) / fY(0.5)
For 1 < x < 3/2:
To find fX|Y(x | 0.5) in this range, divide the joint PDF by fY(0.5):
fX|Y(x | 0.5) = fXY(x, 0.5) / fY(0.5)
To evaluate E[R | A], we need to find the conditional expectation of R given the event A, which is defined as X < 0.5. This can be done using the conditional probability formula:
E[R | A] = integral (over the range of R) of r * P(R=r | A) dr
Since R = XY, we can write:
E[R | A] = integral (over the range of XY) of xy * P(XY = xy | X < 0.5) d(xy)
Now, we need to calculate the conditional probability P(XY = xy | X < 0.5) by using the joint PDF and fact that X < 0.5.
Once we calculate the conditional probability, we can substitute it into the formula and calculate E[R | A]. We integrate over the range of XY by considering the limits of X and Y based on the given ranges.
Note: The calculations for each step may require additional information, such as the specific form of the joint PDF.