Let X be a non-negative random variable. Find the PDF of the random variable Y=lnX for each of the following cases:

For general fX, fY(y)=

Correct answer: fX(e^y)*e^y

2. When fX(x) = {1/4, if 2<x≤6,
0, otherwise,

we have fY(y) = {g(y), if a<y≤b,
0, otherwise.

Correct answer: g(y) = e^(y)/4
a= 0.6931
b= 1.7917

3. When fX(x) = {2(x−1), if 1<x≤2,
0, otherwise,

we have fY(y) = {g(y), if a<y≤b,
0, otherwise.

Correct answer: g(y)= 2*(e^y -1)*(e^(y))
a= 0
b= 0.6931

Well, the actual answers are not really funny, but let's see if we can add a touch of humor to them!

For case 2, it seems like our friend Y is on a logarithmic adventure. With fX(x) = {1/4, if 2<x≤6, and 0 otherwise, the PDF of Y, fY(y), can be expressed as g(y) within a certain range of a to b. And the magical equation is... g(y) = e^(y)/4! Yay, that's how Y rolls!

Now, onto case 3! Our pal Y is exploring an exponential journey this time. With fX(x) = {2(x−1), if 1<x≤2, and 0 otherwise, we have a special function g(y) that governs Y within the range of a to b. Brace yourselves, because g(y) = 2*(e^y - 1)*(e^(y))! Who knew our friend Y had such an exponential sense of humor?

And there you have it! The PDFs of Y for both cases, expressed in their full mathematical glory. Keep on crunching those numbers, and remember to add a dash of humor whenever possible!

To find the PDF (probability density function) of the random variable Y=lnX in each of the given cases, we will use the transformation method.

Case 1:
When the general PDF of X is given as fX(x), the PDF of Y, denoted as fY(y), can be found using the following formula:

fY(y) = fX(e^y) * | d(e^y) / dy |

Since X is a non-negative random variable and Y=lnX, the support for Y will be from negative infinity to infinity.

Therefore, the correct answer is fY(y) = fX(e^y) * e^y.

Case 2:
In this case, the PDF of X is given as follows:

fX(x) = { 1/4, if 2 < x ≤ 6,
0, otherwise.

To find the PDF of Y, we need to determine the range of values of Y and the function g(y) for which fY(y) is non-zero.

Since Y=lnX, we know that X = e^Y. So, the range of Y will be from ln2 to ln6.

Next, we need to find the function g(y) when a < y ≤ b, where a and b are the endpoints of the range of Y.

To find g(y), we substitute X = e^Y into fX(x) and solve for g(y):

g(y) = fX(e^y) = { 1/4, if 2 < e^y ≤ 6,
0, otherwise.

Since g(y) is constant within the range a < y ≤ b, we can take its value as the constant value for that interval.

Therefore, the correct answer is g(y) = e^y / 4 and the range is from a = ln2 ≈ 0.6931 to b = ln6 ≈ 1.7917.

Case 3:
In this case, the PDF of X is given as follows:

fX(x) = { 2(x-1), if 1 < x ≤ 2,
0, otherwise.

Using the same approach as in Case 2, we find that the range of Y is from ln1 = 0 to ln2 = 0.6931.

Next, we find g(y) by substituting X = e^Y into fX(x):

g(y) = fX(e^y) = 2(e^y - 1) * (e^y).

Since g(y) is not a constant within the range a < y ≤ b, we cannot simplify it further.

Therefore, the correct answer is g(y) = 2 * (e^y - 1) * (e^y), and the range is from a = 0 to b = ln2 ≈ 0.6931.

To find the PDF of the random variable Y=lnX, we need to use the transformation of random variables method. Here's how to derive the PDF in each of the given cases:

1. For general fX:
The first step is to find the cumulative distribution function (CDF) of Y, denoted as FY(y):
FY(y) = P(Y ≤ y) = P(lnX ≤ y) = P(X ≤ e^y) = FX(e^y),
where FX(x) is the CDF of X.

To get the PDF of Y, we differentiate the CDF with respect to y:
fY(y) = d/dy [FX(e^y)] = fX(e^y) * d(e^y)/dy = fX(e^y) * e^y.

Therefore, for general fX, the PDF of Y is fY(y) = fX(e^y) * e^y.

2. For the given fX(x):
We know that fX(x) = 1/4 for 2<x≤6, and 0 otherwise.

To find the range of Y, we need to find the values of a and b such that a<y≤b. Rearranging the equation Y=lnX, we have X=e^Y. Therefore, a=e^a and b=e^b.

To find g(y), the PDF of Y within the range (a,b), we calculate:
g(y) = fX(e^y) = 1/4, for e^a<y≤e^b, and 0 otherwise.

Using e^a=0.6931 and e^b=1.7917, we get:
g(y) = 1/4, for 0.6931<y≤1.7917, and 0 otherwise.

Therefore, the PDF of Y is fY(y) = g(y) = e^y/4, for 0.6931<y≤1.7917, and 0 otherwise.

3. For the given fX(x):
We know that fX(x) = 2(x−1) for 1<x≤2, and 0 otherwise.

Again, we need to find the range of Y by solving X=e^Y, which gives us a=0 and b=0.6931.

To find g(y), the PDF of Y within the range (a,b), we calculate:
g(y) = fX(e^y) = 2(e^y - 1) * (d(e^y)/dy) = 2(e^y - 1) * e^y.

Therefore, the PDF of Y is fY(y) = g(y) = 2*(e^y - 1)*(e^y), for 0<y≤0.6931, and 0 otherwise.

In summary, to find the PDF of the random variable Y=lnX, we use the transformation of random variables method, derive the CDF of Y, differentiate it to get the PDF, and consider the range of Y based on the given fX(x) and the equation Y=lnX.