Paul is vacationing in Monte Carlo. On any given night, he 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 2x.
Determine the joint PDF fX,Y(x,y).
If 0<x<40 and 0<y<2x,
fX,Y(x,y)=
MY ANSWER : 1/1600
If y<0 or y>2x,
fX,Y(x,y)=
MY ANSWER:0
On any particular night, Paul makes a profit of Z=Y−X dollars. Find the probability that Paul makes a positive profit (i.e., P(Z>0)):
MY ANSWER :0.5
Find the PDF of Z. Express your answers in terms of z using standard notation . Hint: Start by finding fZ|X(z|x).
If 0<z<40, fZ(z)=
couldn't answer
If −40<z<0, fZ(z)=
couldn't answer
If z<−40 or z>40, fZ(z)=
couldn't answer
What is E[Z]?
E[Z]=
couldn't answer
what if we had 3x instead of 2x
To determine the joint PDF fX,Y(x,y), we need to consider the given information that conditional on X=x, Y is uniformly distributed between zero and 2x.
For 0 < x < 40 and 0 < y < 2x, the joint PDF is given by:
fX,Y(x, y) = fX(x) * fY|X(y|x)
fX(x) is the PDF of X, as shown in the figure. Since it is not specified in the question or figure, we cannot determine its specific form.
fY|X(y|x) is the conditional PDF of Y given X=x. In this case, Y is uniformly distributed between zero and 2x. The probability density function for a uniform distribution on the interval [a, b] is given by fY(y) = 1 / (b - a).
Therefore, for 0 < x < 40 and 0 < y < 2x, the joint PDF is:
fX,Y(x, y) = fX(x) * (1 / (2x - 0)) = fX(x) / (2x)
To find the probability that Paul makes a positive profit (P(Z > 0)), we need to consider that Z = Y - X.
Since Y is uniformly distributed between zero and 2x, the range of possible values for Z is from -x to x. Any value of Z greater than 0 indicates a positive profit.
The probability that Paul makes a positive profit is therefore the probability that Z is in the range (0, x).
P(Z > 0) = P(0 < Z < x) = P(-x < Y - X < x) = P(2x - x < Y < 2x + x) = P(x < Y < 3x)
Since Y is uniformly distributed between zero and 2x, the probability that x < Y < 3x is (3x - x) / (2x - 0) = 2x / (2x) = 1.
So, the probability that Paul makes a positive profit is 1.
To find the PDF of Z, fZ(z), we need to find fZ|X(z|x), the conditional PDF of Z given X=x.
Z = Y - X
Rearranging the equation, Y = Z + X
Since Y is a continuous random variable uniformly distributed between zero and 2x, the conditional PDF of Y given X=x is given by fY|X(y|x) = 1 / (2x - 0) = 1 / (2x).
Substituting Y = Z + X, we have Z + X = y.
Solving for Z, Z = y - X.
Now we can express Z in terms of x and y:
Z = y - X.
To find fZ(z), we need to express Z in terms of z:
Z = y - X = y - (z + X) = y - z - X.
Rearranging the equation, X = y - z - Z.
Substituting X = y - z - Z in the conditional PDF, we have:
fZ|X(z|x) = fY|X(y|x) = 1 / (2x).
Now, we need to find the range of values of z for which fZ(z) is non-zero.
For -40 < z < -20 or 20 < z < 40, there is no overlap between the range of values of z and the joint PDF fX,Y(x, y) (since the domain for x and y is 0 to 40).
Therefore, fZ(z) = 0 for z < -40 or z > 40.
For -20 < z < 0, we have 0 < X < 40 and 0 < Y < 2X, which gives 20 < Y - X < 40.
Therefore, fZ(z) = fY - X(y - x) = 1 / (2x) = 1 / (2(y-z)).
For 0 < z < 20, we have 0 < X < 40 and 0 < Y < 2X, which gives 0 < Y - X < 20.
Therefore, fZ(z) = fY - X(y - z) = 1 / (2x) = 1 / (2(y-z)).
To find the PDF of Z, fZ(z), we need to integrate fZ|X(z|x) with respect to the PDF fX(x) over the range of possible values of x.
Since the specific form of fX(x) is not given in the question or figure, we cannot determine the integral or the PDF of Z, fZ(z).
Therefore, we cannot determine the PDF of Z or find E[Z] without the specific form of fX(x).
To determine the joint PDF fX,Y(x,y), we need to consider the given conditions. Since X has a PDF shown in the figure, we can use that information to find fX(x). Since the continuous random variable Y is uniformly distributed between zero and 2x, we know that for a given x, the probability of Y falling within the range (0, 2x) is 1. Therefore, the joint PDF fX,Y(x,y) can be written as:
fX,Y(x,y) = fX(x) * fY|X(y|x)
Since Y is uniformly distributed between zero and 2x, the conditional PDF fY|X(y|x) is equal to 1/(2x) for y ∈ (0, 2x) and 0 otherwise. We can substitute this into the joint PDF equation:
fX,Y(x,y) = fX(x) * fY|X(y|x)
= fX(x) * (1/(2x)) if y ∈ (0, 2x)
= 0 otherwise
Since we are given the PDF fX(x) in the figure, we can substitute that in to get the final joint PDF expression.
Now, let's move on to finding the probability that Paul makes a positive profit, P(Z>0).
Since Z = Y - X, we can write it as Z = (2X - X) = X.
To find the probability that Paul makes a positive profit, we need to find the probability that X > 0. We have the given condition that 0 < x < 40, so the range of X for which Paul makes a positive profit is (0, 40).
Thus, the probability that Paul makes a positive profit is equal to the probability that X falls within the range (0, 40), which can be calculated using the PDF fX(x) given in the figure.
Next, we need to find the PDF of Z, denoted as fZ(z).
To obtain fZ(z), we need to find fZ|X(z|x) first.
Since Z = Y - X, we can rearrange it as Y = Z + X.
Given that Y is uniformly distributed between zero and 2x, and X is fixed, the conditional PDF of Z given X, fZ|X(z|x), can be obtained by replacing Y by its expression in terms of Z and X:
fZ|X(z|x) = fY|X(z + x|x)
We know that for a given x, the probability of Y falling within the range (0, 2x) is 1 and 0 otherwise. Replacing Y by Z + X, we have:
fZ|X(z|x) = 1 if z + x ∈ (0, 2x)
0 otherwise
To find the PDF of Z, we need to marginalize fZ|X(z|x) over the range of x. However, the expression for fZ|X(z|x) depends on z and x.
Since we don't have the joint PDF of X and Z, we cannot calculate fZ(z) without additional information.
Similarly, to find E[Z], we need the PDF of Z, fZ(z). Without the PDF of Z, we cannot calculate the expected value E[Z].
Unfortunately, based on the given information in the question, we cannot answer the last two parts related to the PDF of Z and E[Z]. We would need more information about the joint distribution of X and Z to solve those parts.