Let X and Y be independent positive random variables. Let Z=X/Y . In what follows, all occurrences of x , y , z are assumed to be positive numbers.
1) Suppose that X and Y are discrete, with known PMFs, pX and pY . Then,
pZ|Y(z|y)=pX(?).
What is the argument in the place of the question mark?
----------------------?
2) Suppose that X and Y are continuous, with known PDFs, fX and fY . Provide a formula, analogous to the one in part (a), for fZ|Y(z|y) in terms of fX . That is, find A and B in the formula below.
fZ|Y(z|y)=AfX(B).
A= ?
B= ?
Which of the following is a formula for fZ(z) ?
fZ(z)=…
(Choose all that apply.)
a) fZ(z)=∫∞0fY,Z(y,z)dy
b) fZ(z)=∫∞0fY,Z(y,z)dz
c) fZ(z)=∫∞0fY(y)fZ,Y(z,y)dy
d) fZ(z)=∫∞0fY(y)fZ|Y(z|y)dy
e) fZ(z)=∫∞0fY(y)fX(yz)dy
f) fZ(z)=∫∞0yfY(y)fX(yz)dy
Check the last option in 3). You can find all answers here.
a,d,f
I am still lost here. Can anyone give me a hint?
www.jiskha.com/questions/1822154/let-and-be-independent-positive-random-variables-let-in-what
1) The argument in the place of the question mark is X/z.
2) A = z and B = y/z.
The formula for fZ(z) is:
fZ(z) = ∫∞0 fY(y) fX(yz) dy
So, option e) fZ(z) = ∫∞0 fY(y) fX(yz) dy is the correct formula for fZ(z).
To answer the questions, let's break down the problems one by one:
1) When X and Y are discrete random variables, the formula for conditional probability mass function (PMF) is given by:
pZ|Y(z|y) = pX(X)
Here, pX(X) is the PMF of X. In this case, X and Y being independent implies that their probability distributions are unrelated. Therefore, the argument in the place of the question mark would be the value of X.
2) When X and Y are continuous random variables, the formula for conditional probability density function (PDF) is given by:
fZ|Y(z|y) = A * fX(B)
To find A and B, we need to consider how Z is derived from X and Y. Since Z = X/Y, we can rewrite it as Z * Y = X.
Therefore, A = 1/y (because Z = X/Y) and B = yz (because Z * Y = X).
Now, let's determine the formula for fZ(z) among the given options:
a) fZ(z) = ∫∞0fY,Z(y,z)dy
This option suggests integrating the joint PDF of Y and Z with respect to Y. The answer would be the integral of the joint PDF.
b) fZ(z) = ∫∞0fY,Z(y,z)dz
This option suggests integrating the joint PDF of Y and Z with respect to Z. This is not the correct formula.
c) fZ(z) = ∫∞0fY(y)fZ,Y(z,y)dy
This option suggests multiplying the PDF of Y with the conditional PDF of Z given Y, then integrating with respect to Y. This is not the correct formula.
d) fZ(z) = ∫∞0fY(y)fZ|Y(z|y)dy
This option suggests multiplying the PDF of Y with the conditional PDF of Z given Y, then integrating with respect to Y. This is the correct formula.
e) fZ(z) = ∫∞0fY(y)fX(yz)dy
This option suggests multiplying the PDFs of Y and X, where Y is multiplied by the argument z, then integrating with respect to Y. This is not the correct formula.
f) fZ(z) = ∫∞0yfY(y)fX(yz)dy
This option suggests multiplying the PDFs of Y and X, where Y is multiplied by the argument z and integrated with respect to Y. This is the correct formula.
Therefore, the formulas for fZ(z) are d) fZ(z) = ∫∞0fY(y)fZ|Y(z|y)dy and f) fZ(z) = ∫∞0yfY(y)fX(yz)dy.