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# 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

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1. 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

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2. 👎
3. ℹ️
4. 🚩
2. 4. Find fX∣Y(x∣0.5).
If 1/2<x<1:
fX|Y(x∣0.5)=0.5

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2. 👎
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3. 4. Find fX(x).
If 1<x<2:

fX(x) = (-3)*x/2+3

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2. 👎
3. ℹ️
4. 🚩