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 X has the PDF shown in the figure. Conditional on X=x , the continuous random variable Y is uniformly distributed between zero and 3x
1. Determine the joint PDF fX,Y(x,y).
If 0<x<40 and 0<y<3x:
fX,Y(x,y)= ?
If y<0 or y>3x:
fX,Y(x,y)= ?
2. On any particular night, Sophia makes a profit Z=Y−X dollars. Find the probability that Sophia makes a positive profit, that is, find P(Z>0).
P(Z>0)= ?
3. Find the PDF of Z. Express your answers in terms of z using standard notation.
Hint: Start by finding fZ∣X(z∣x).
If −40<z<0:
fZ(z)= ?
If 0<z<80:
fZ(z)= ?
If z<−40 or z>80:
fZ(z)= ?
4. What is E[Z]?
E[Z]= ?
1.a. 1/2400
1.b. 0
2. 2/3
3.c. 0 (zero)
The others I don't know
What figure? Cannot copy and paste here.
Show how prime numbers can be multiplied to equal 27. Help
hi i need help in sience
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1. To determine the joint PDF fX,Y(x,y), we need to consider the given conditions.
For 0 < x < 40 and 0 < y < 3x:
Since X is given as a continuous random variable with a probability density function (PDF) shown in the figure, we can write the PDF for X as:
fX(x) = 1/40 for 0 < x < 40
Given that Y is uniformly distributed between zero and 3x given X = x, we can write the conditional PDF for Y as:
fY|X(y|x) = 1/(3x) for 0 < y < 3x, and 0 otherwise
The joint PDF fX,Y(x,y) is obtained by multiplying the PDFs for X and Y:
fX,Y(x,y) = fX(x) * fY|X(y|x)
For 0 < x < 40 and 0 < y < 3x:
fX,Y(x,y) = (1/40) * (1/(3x))
If y < 0 or y > 3x:
Since Y is uniformly distributed between zero and 3x, the PDF is zero outside this range. So:
fX,Y(x,y) = 0
2. To find the probability that Sophia makes a positive profit, P(Z > 0), we need to consider the difference between Y and X (Z = Y - X).
P(Z > 0) = P(Y - X > 0)
We can rewrite this as:
P(Y > X) = ∫∫ fX,Y(x,y) dy dx, where the limits of integration are 0 < x < 40 and x < y < 3x.
Substituting the previously derived expression for fX,Y(x,y):
P(Z > 0) = ∫∫ fX,Y(x,y) dy dx
= ∫∫ (1/40) * (1/(3x)) dy dx, where the limits of integration are 0 < x < 40 and x < y < 3x.
Solving this double integral will give you the probability that Sophia makes a positive profit.
3. To find the PDF of Z, fZ(z), we need to find the conditional PDF of Z given X = x, denoted as fZ|X(z | x).
Z = Y - X
Given that Y is uniformly distributed between zero and 3x, we can express Y as:
Y = Z + X
Rearranging the equation, we can find the inverse:
X = Y - Z
Taking the derivative of X with respect to Z, we get:
dX/dZ = -1
By applying the change of variables formula for continuous random variables, we have:
fZ|X(z | x) = fY|X(z + x | x) * |dX/dZ|
Since Y is uniformly distributed between zero and 3x, the conditional PDF of Y given X = x is:
fY|X(y | x) = 1/(3x) for 0 < y < 3x, and 0 otherwise
Plugging these values into the equation, we get:
fZ|X(z | x) = (1/(3x)) * |dX/dZ|
= (1/(3x)) * 1
Simplifying, we get:
fZ|X(z | x) = 1/(3x) for z + x > 0 and z + x < 3x, and 0 otherwise
To find the PDF of Z, we need to marginalize over X to remove the conditioning. Integrating fZ|X(z | x) with respect to x will give us fZ(z):
fZ(z) = ∫ fZ|X(z | x) fX(x) dx, where the limits of integration depend on z.
For -40 < z < 0:
To find fZ(z), we need to integrate fZ|X(z | x) * fX(x) over the relevant range for z and x.
For -40 < z < 0, the integration limits will depend on z and result in the following expression for fZ(z).
fZ(z) = ∫ (1/(3x))(1/40) dx
For 0 < z < 80:
To find fZ(z), again we need to integrate fZ|X(z | x) * fX(x) over the relevant range for z and x.
For 0 < z < 80, the integration limits will depend on z and result in the following expression for fZ(z).
fZ(z) = ∫ (1/(3x))(1/40) dx
If z < -40 or z > 80:
Outside the range of -40 < z < 80, the PDF fZ(z) will be zero.
4. To find the expected value of Z, denoted as E[Z], we need to calculate the integral of z * fZ(z) over the entire support of Z.
E[Z] = ∫ z * fZ(z) dz, over the appropriate range for Z.
Substituting the previously derived expressions for fZ(z), we can calculate the expected value of Z by evaluating this integral.