The random variables X and Y are jointly continuous, with a joint PDF of the form
fX,Y(x,y)=
{cxy if 0≤x≤y≤1
0 otherwise}
where c is a normalizing constant.
a) Is it true that fX|Y(2|0.5) is equal to zero?
b) Is it true that fX|Y(0.5|2) is equal to zero?
a) yes
b) no
a) Well, my friend, in order to determine fX|Y(2|0.5), we need to focus on the range of values for X and Y. Given that 0≤x≤y≤1, we can see that X=2 doesn't fall within this range. Therefore, it is safe to say that fX|Y(2|0.5) is indeed equal to zero. Zero chance of existence!
b) Ah, the plot thickens! Looking at fX|Y(0.5|2), we find ourselves in a conundrum. Again, considering the range 0≤x≤y≤1, we see that Y=2 is outside this range. So, once more, we conclude that fX|Y(0.5|2) is equal to zero. Zero humor, zero probability.
a) To determine the conditional PDF fX|Y(2|0.5), we need to find the value of c that normalizes the joint PDF.
First, let's find the range of values for x and y that satisfy the condition 0 ≤ x ≤ y ≤ 1 and determine the area over which the joint PDF is non-zero.
Given that 0 ≤ x ≤ y ≤ 1, the condition x = 2 with y = 0.5 does not satisfy this inequality and falls outside the range of the joint PDF. Therefore, fX|Y(2|0.5) is equal to zero.
b) Similarly, using the same logic, when x = 0.5 and y = 2, it does not satisfy the condition 0 ≤ x ≤ y ≤ 1. Hence, fX|Y(0.5|2) is also equal to zero.
Therefore, both statements are true.
To determine whether fX|Y(2|0.5) and fX|Y(0.5|2) are equal to zero, we need to calculate the conditional PDFs (probability density functions).
Let's start with fX|Y(2|0.5):
The conditional PDF of X given Y is denoted as fX|Y(x|y) and is calculated as the joint PDF divided by the marginal PDF of Y. In this case:
fX|Y(x|y) = fX,Y(x,y) / fY(y)
To find fY(y), we integrate the joint PDF with respect to x over the range of possible x values:
fY(y) = ∫[from 0 to y] cxy dx
Now let's substitute the given values into the formulas:
fX|Y(2|0.5) = fX,Y(2,0.5) / fY(0.5)
The joint PDF fX,Y(x,y) is given as cxy, so plugging in the values:
fX,Y(2,0.5) = c * 2 * 0.5
To find fY(0.5), we integrate the joint PDF over the range of x values, from 0 to y=0.5:
fY(0.5) = ∫[from 0 to 0.5] cxy dx = c * ∫[from 0 to 0.5] xy dx
Now we can substitute the known values into the equation and calculate the integral:
fY(0.5) = c * ∫[from 0 to 0.5] xy dx = c * [0.5 * (0.5)^2] = 0.125c
Finally, we can calculate fX|Y(2|0.5):
fX|Y(2|0.5) = fX,Y(2,0.5) / fY(0.5) = (c * 2 * 0.5) / (0.125c) = 4
Therefore, fX|Y(2|0.5) is equal to 4, not zero.
Now let's move on to fX|Y(0.5|2):
Using the same approach, we have:
fX|Y(0.5|2) = fX,Y(0.5,2) / fY(2)
The joint PDF fX,Y(x,y) is given as cxy, so plugging in the values:
fX,Y(0.5,2) = c * 0.5 * 2 = c
To find fY(2), we integrate the joint PDF over the range of x values, from 0 to y=2:
fY(2) = ∫[from 0 to 2] cxy dx = c * ∫[from 0 to 2] 2x dx = c * (2 * (2^2)/2) = 4c
Now we can calculate fX|Y(0.5|2):
fX|Y(0.5|2) = fX,Y(0.5,2) / fY(2) = c / (4c) = 1/4
Therefore, fX|Y(0.5|2) is equal to 1/4, not zero.
To summarize,
a) fX|Y(2|0.5) is equal to 4, not zero.
b) fX|Y(0.5|2) is equal to 1/4, not zero.