Question

show that f(x0=3/x^4 is a valid PDF where x+>1

Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

The pdf of Y is fY(y)=2y3.

Step-by-step explanation:

The first thing to do in such a problem is to figure out the support of the random variable YY, which is (1,∞).(1,∞). To find it, try several values of xx in (0,1)(0,1) and see what happens. You were right to ask about this crucial issue.

Your text probably shows transformation methods based on the CDF and on the PDF. An abbreviated version of the former (with gaps for you to complete) is as follows:

FY(t)=P(Y≤t)=P(1/X≤t)=P(X≥1/t)=1−P(X≤1/t)=1−1/t2,FY(t)=P(Y≤t)=P(1/X≤t)=P(X≥1/t)=1−P(X≤1/t)=1−1/t2,

for tt in the support of YY.

Then, by differentiation, the PDF of YY is fY(t)=2t−3,fY(t)=2t−3, as you say. Finally, E(Y)=∫∞12t−3dt=2.E(Y)=∫1∞2t−3dt=2. [Some of this is in your Question, and some is in @drhab's succinct Answer (+1)]

If you recognize that X∼Beta(2,1),X∼Beta(2,1), then it is easy to do a quick simulation in R to check this answer (correct to two or three decimal places based on a million simulated values of 

Below is a histogram of 10,000 simulated values of YY along with a plot of the PDF of YY. This distribution is extremely right-skewed, with a long tail to the right. The scale of the histogram extends out to about 60 in recognition of occasional scattered points too sparsely spread to make histogram bars of noticeable height. (The biggest of the 10,000 values plotted in this histogram was at 57.44494.)

Answer 2

Answer:

The first thing to do in such a problem is to figure out the support of the random variable YY, which is (1,∞).(1,∞). To find it, try several values of xx in (0,1)(0,1) and see what happens. You were right to ask about this crucial issue.

Your text probably shows transformation methods based on the CDF and on the PDF. An abbreviated version of the former (with gaps for you to complete) is as follows:

FY(t)=P(Y≤t)=P(1/X≤t)=P(X≥1/t)=1−P(X≤1/t)=1−1/t2,FY(t)=P(Y≤t)=P(1/X≤t)=P(X≥1/t)=1−P(X≤1/t)=1−1/t2,

for tt in the support of YY.

Then, by differentiation, the PDF of YY is fY(t)=2t−3,fY(t)=2t−3, as you say. Finally, E(Y)=∫∞12t−3dt=2.E(Y)=∫1∞2t−3dt=2. [Some of this is in your Question, and some is in @drhab's succinct Answer (+1)]

If you recognize that X∼Beta(2,1),X∼Beta(2,1), then it is easy to do a quick simulation in R to check this answer (correct to two or three decimal places based on a million simulated values of

Below is a histogram of 10,000 simulated values of YY along with a plot of the PDF of YY. This distribution is extremely right-skewed, with a long tail to the right. The scale of the histogram extends out to about 60 in recognition of occasional scattered points too sparsely spread to make histogram bars of noticeable height. (The biggest of the 10,000 values plotted in this histogram was at 57.44494.)