From what I gather from your latest comments, the PDF is given to be
and in particular, <em>f(x, y)</em> = <em>cxy</em> over the unit square [0, 1]², meaning for 0 ≤ <em>x</em> ≤ 1 and 0 ≤ <em>y</em> ≤ 1. (As opposed to the unbounded domain, <em>x</em> ≤ 0 *and* <em>y</em> ≤ 1.)
(a) Find <em>c</em> such that <em>f</em> is a proper density function. This would require
(b) Get the marginal density of <em>X</em> by integrating the joint density with respect to <em>y</em> :
(c) Get the marginal density of <em>Y</em> by integrating with respect to <em>x</em> instead:
(d) The conditional distribution of <em>X</em> given <em>Y</em> can obtained by dividing the joint density by the marginal density of <em>Y</em> (which follows directly from the definition of conditional probability):
(e) From the definition of expectation:
(f) Note that the density of <em>X</em> | <em>Y</em> in part (d) identical to the marginal density of <em>X</em> found in (b), so yes, <em>X</em> and <em>Y</em> are indeed independent.
The result in (e) agrees with this conclusion, since E[<em>XY</em>] = E[<em>X</em>] E[<em>Y</em>] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)
Answer:
144 cups-14 pints
Step-by-step explanation:
5.75+ 0.75x is your answer because the only thing that will change is the number of pounds
Answer:
amount of your paycheck is the correct answer because the summation of all check paid to me will be my net earning.
Answer:
A. ?
B. ?
C. ?
1. Different
2. Same
3. Not always
4. multiplying, same, different
5. Don't know
6. and 7. don't have any students or anything so I can't answer that
Step-by-step explanation:
I kinda went off what I knew
Hey at least I tried