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. Find the PDF of Z . Express your answers in terms of z using standard notation.
If −40<z<0 :
fZ(z)=
If 0<z<80 :
fZ(z)=
If z<−40 or z>80 :
fZ(z)=
2. What is E[Z] ?
E[Z]=
3.
c. 0
4. 40/3
1:
1/2400
0
2:
2/3
3:
(40+z)/2400
no answer
0
4:
no answer
5/7 result
3.
b) (-z/4800)+(1/60)
1. 1/2400, 0
2. 2/3
3. (40+z)/2400, (-z/4800)+(1/60), 0
4. 40/3
1. Since Y is uniformly distributed between 0 and 3X, the probability density function (PDF) of Y given X = x is 1/(3x). We can use this information to find the PDF of Z.
If −40 < z < 0:
Since Z = Y - X, we need to find the probability that Z is in this range. This occurs when the difference between Y and X is negative, so we have:
fZ(z) = ∫[0,∞] (1/(3x)) * dx, where x = (z + y).
Simplifying this integral, we get:
fZ(z) = (1/3) * ∫[0,∞] (1/x) * dx
= (1/3) * ln(x) |[0,∞]
= (1/3) * (ln(∞) - ln(0))
= (1/3) * (∞ - (-∞))
= (1/3) * (∞ + ∞)
= (1/3) * ∞
= ∞
If 0 < z < 80:
Similarly, for this range, we need to find the probability that Z is between z and z + 3x. This occurs when Y - X is between 0 and z.
fZ(z) = ∫[0,∞] (1/(3x)) * dx, where x = (z + y)/3.
Simplifying this integral, we get:
fZ(z) = (1/3) * ∫[0,∞] (1/x) * dx
= (1/3) * ln(x) |[0,∞]
= (1/3) * (ln(∞) - ln(0))
= (1/3) * (∞ - (-∞))
= (1/3) * (∞ + ∞)
= (1/3) * ∞
= ∞
If z < −40 or z > 80:
For these ranges, the probability density function is 0 since Z cannot take on values outside of the range [-40, 80]. So:
fZ(z) = 0
2. To find E[Z], we need to find the expected value of Z, i.e., the average value of Z.
E[Z] = ∫[-40,0] z * fZ(z) dz + ∫[0,80] z * fZ(z) dz
(integral from -40 to 0 is 0 because fZ(z) = 0 for z < -40)
= ∫[0,80] z * ∞ dz
= ∞ * ∫[0,80] z dz
= ∞ * (z^2/2) |[0,80]
= ∞ * (80^2/2)
= ∞ * 3200
= ∞
So, E[Z] = ∞.
To find the probability density function (PDF) of Z, we need to use the properties of conditional probability and the cumulative distribution function (CDF). Let's break down the steps:
1. Find the PDF of Z:
First, note that Z is derived from X and Y:
Z = Y - X
To calculate fZ(z) (PDF of Z), we need to consider three cases:
Case 1: When -40 < z < 0:
In this case, we have Z < 0. Therefore, we need to calculate the probability that Z is less than z.
Using conditional probability, we get:
P(Z < z) = P(Y - X < z | X = x) = P(Y < z + x | X = x)
Since Y is uniformly distributed between 0 and 3x, the CDF of Y is given by:
FY(y | X = x) = (y - 0) / (3x - 0) = y / (3x)
Thus, we can rewrite the probability as:
P(Y < z + x | X = x) = (z + x) / (3x)
To obtain the PDF, we differentiate the CDF with respect to z:
fZ(z) = d/dz [(z + x) / (3x)] = 1 / (3x)
Since -40 < z < 0, we must integrate the PDF over this range:
fZ(z) = ∫[from -40 to 0] (1 / (3x)) dx = (1/3) * ln|3x| [from -40 to 0]
= (1/3) * (ln|-120| - ln|-1200|)
= (1/3) * (ln(1200) - ln(120))
Thus, if -40 < z < 0:
fZ(z) = (1/3) * (ln(1200) - ln(120))
Case 2: When 0 < z < 80:
In this case, we have Z > 0. Therefore, we need to calculate the probability that Z is greater than z.
Using conditional probability, we get:
P(Z > z) = P(Y - X > z | X = x) = P(Y > z + x | X = x)
Again, using the CDF of Y, we have:
P(Y > z + x | X = x) = (3x - (z + x)) / (3x) = (2x - z) / (3x)
Differentiating the CDF with respect to z to obtain the PDF:
fZ(z) = d/dz [(2x - z) / (3x)] = -1 / (3x)
Since 0 < z < 80, we integrate the PDF over this range:
fZ(z) = ∫[from 0 to 80] (-1 / (3x)) dx = -(1/3) * ln|3x| [from 0 to 80]
= -(1/3) * (ln(240) - ln(0))
Note: ln(0) is undefined, but it will cancel out with the lower limit of integration.
Thus, if 0 < z < 80:
fZ(z) = -(1/3) * ln(240)
Case 3: When z < -40 or z > 80:
In these cases, Z is outside the range [-40, 80], meaning it has zero probability. Hence, fZ(z) = 0.
Therefore, the final PDF of Z is:
If -40 < z < 0:
fZ(z) = (1/3) * (ln(1200) - ln(120))
If 0 < z < 80:
fZ(z) = -(1/3) * ln(240)
If z < -40 or z > 80:
fZ(z) = 0
2. To find E[Z] (the expected value of Z), we integrate Z multiplied by its PDF, fZ(z), over the entire range of Z:
E[Z] = ∫[from -∞ to ∞] z * fZ(z) dz
In this case, since the PDF is zero for z < -40 and z > 80, the range of the integral becomes -40 to 80.
E[Z] = ∫[from -40 to 80] z * fZ(z) dz
Using our previous derived PDF expressions, we can substitute them into the integral and calculate E[Z].