# Thread: Wondering if I did these correctly

1. ## 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.

2. 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 $\displaystyle \frac{1}{2} \int_2^8 (y-2) \, dy = 1$ ....?

2) Does $\displaystyle 2 \int_{-\infty}^{+ \infty} e^{-y^2} \, dy = 1$ ....?

3. 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]$\displaystyle \in R$.
Y is another set of random number [2,8] $\displaystyle \in R$ where Y(X).
The total area under your f(x) and f(y), [a,b]$\displaystyle \in R$, both, must equal to 1, i.e, the integration of f(x) and f(y) [a,b]$\displaystyle \in R$ must equal to 1, regardless, but first make sure that f(y) is correct before you go too far.