Thread: Wondering if I did these correctly

Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

1) Suppose the p.d.f. of X is

f(x) = 1/2x for 0 < x < 2
and 0 otherwise.

Determine the p.d.f. of Y = 3X + 2.

My solution:

I found the inverse of Y = 3X + 2.
This came out to be (y-2)/3 = x.

Now, I substituted this value in for x in f(x).
This gave f(x) = (3/2)/(y-2).

Then, I found the derivative of (y-2)/3.
This turned out to be 1/3.

Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

So if the above work is correct, the p.d.f. of Y = 3X + 2 is

(1/2)(y-2) for 2 < y < 8
and 0 otherwise.

2) Suppose the p.d.f. of X is

f(x) = e^-x for x > 0,
and 0 for x less than or equal to 0.

Determine the p.d.f. of Y = X^(1/2)

I found the inverse of Y = X^(1/2).
This turned out to be Y^2 = X.
Then, I subbed in y^2 for x in f(x).
This gave e^(-y^2).
Now, the derivative of y^2 is 2.
Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.

Originally Posted by Integrator020

Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

1) Suppose the p.d.f. of X is

f(x) = 1/2x for 0 < x < 2
and 0 otherwise.

Determine the p.d.f. of Y = 3X + 2.

My solution:

I found the inverse of Y = 3X + 2.
This came out to be (y-2)/3 = x.

Now, I substituted this value in for x in f(x).
This gave f(x) = (3/2)/(y-2).

Then, I found the derivative of (y-2)/3.
This turned out to be 1/3.

Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

So if the above work is correct, the p.d.f. of Y = 3X + 2 is

(1/2)(y-2) for 2 < y < 8
and 0 otherwise.

2) Suppose the p.d.f. of X is

f(x) = e^-x for x > 0,
and 0 for x less than or equal to 0.

Determine the p.d.f. of Y = X^(1/2)

I found the inverse of Y = X^(1/2).
This turned out to be Y^2 = X.
Then, I subbed in y^2 for x in f(x).
This gave e^(-y^2).
Now, the derivative of y^2 is 2.
Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.

1) Does ....?

2) Does ....?

Originally Posted by Integrator020

Hello everyone! I solved (or rather, tried to) a couple of problems, and I was wondering if I attempted them correctly. Any feedback would be wonderful! Thank you in advance!

1) Suppose the p.d.f. of X is

f(x) = 1/2x for 0 < x < 2
and 0 otherwise.

Determine the p.d.f. of Y = 3X + 2.

My solution:

I found the inverse of Y = 3X + 2.
This came out to be (y-2)/3 = x.

Now, I substituted this value in for x in f(x).
This gave f(x) = (3/2)/(y-2).

Then, I found the derivative of (y-2)/3.
This turned out to be 1/3.

Multiplying (1/3) * (3/2)/(y-2) gave me (1/2)(y-2).

So if the above work is correct, the p.d.f. of Y = 3X + 2 is

(1/2)(y-2) for 2 < y < 8
and 0 otherwise.

2) Suppose the p.d.f. of X is

f(x) = e^-x for x > 0,
and 0 for x less than or equal to 0.

Determine the p.d.f. of Y = X^(1/2)

I found the inverse of Y = X^(1/2).
This turned out to be Y^2 = X.
Then, I subbed in y^2 for x in f(x).
This gave e^(-y^2).
Now, the derivative of y^2 is 2.
Multiplying gives 2*e^(-y^2) as the p.d.f., where y lies between negative infinity and positive infinity.

If you can solve the 1st problem, you should be able to solve the 2nd one. The idea is the same.

You are give f(x) as pdf, where x is a set of random numbers in [0,2] .
Y is another set of random number [2,8] where Y(X).
Your f(y) = f(Y(X=x))

You have found you inverse function correctly, but after you substitute it into f(x) you made two mistakes: (1) you should get f(y); (2) you should have two terms in the function.

The derivative that you got was meaning less because it does not say what slope you are looking for. Are you looking for the slope of f(y)? Where do you think the slope will take you? If you don't need it, why look for it?

The total area under your f(x) and f(y), [a,b] , both, must equal to 1, i.e, the integration of f(x) and f(y) [a,b] must equal to 1, regardless, but first make sure that f(y) is correct before you go too far.