Probability Density Function Question

If someone could review my work on this exercise :)

When I use f X(x) is the pdf of X and fY(y) is the pdf of Y.

Consider X a random variable with pdf

f X(x) = (3/8) * (1-x)^2 if -1<=x<=1

and 0 otherwise.

Find the pdf of Y = 1-X^2

**My workings:**

If -1<=x<=1 then 0<=Y<=1

Then P(Y<=y) = P(1-X^2<=y) = P(X^2>=1-y) = P(X>=sqrt(1-y))

=1-P(X<=sqrt(1-y))

=1 - (3/8) * (1 - sqrt(1-y)) ^ 2 for 0<=y<=1

and 0 otherwise

Now f Y (y) = d/dy (P(Y<=y))

= (3/4) * (1-sqrt(1-y)) / (2*sqrt(1-y)) for 0<=y<=1 and 0 otherwise!

**Or should I say P(X^2>=1-y) = 1-P(X^2<=1-y) = 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) and then how do I proceed?**

Re: Probability Density Function Question

Quote:

Originally Posted by

**Darkprince** If someone could review my work on this exercise :)

When I use f X(x) is the pdf of X and fY(y) is the pdf of Y.

Consider X a random variable with pdf

f X(x) = (3/8) * (1-x)^2 if -1<=x<=1

and 0 otherwise.

Find the pdf of Y = 1-X^2

**My workings:**

If -1<=x<=1 then 0<=Y<=1

Then P(Y<=y) = P(1-X^2<=y) = P(X^2>=1-y) = P(X>=sqrt(1-y))

=1-P(X<=sqrt(1-y))

=1 - (3/8) * (1 - sqrt(1-y)) ^ 2 for 0<=y<=1

and 0 otherwise

Now f Y (y) = d/dy (P(Y<=y))

= (3/4) * (1-sqrt(1-y)) / (2*sqrt(1-y)) for 0<=y<=1 and 0 otherwise!

**Or should I say P(X^2>=1-y) = 1-P(X^2<=1-y) = 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) and then how do I proceed?**

$\displaystyle cdf = G(y) = \Pr(Y < y) = \Pr(1 - X^2 < y)$

$\displaystyle = \Pr(X > \sqrt{1 - y}) + \Pr(X < -\sqrt{1 - y}) = .... $

Therefore $\displaystyle pdf = g(y) = \frac{dG}{dy} .....$ for $\displaystyle 0 \leq y \leq 1$ and 0 otherwise.

Re: Probability Density Function Question

Quote:

Originally Posted by

**mr fantastic** $\displaystyle cdf = G(y) = \Pr(Y < y) = \Pr(1 - X^2 < y)$

$\displaystyle = \Pr(X > \sqrt{1 - y}) + \Pr(X < -\sqrt{1 - y}) = .... $

Therefore $\displaystyle pdf = g(y) = \frac{dG}{dy} .....$ for $\displaystyle 0 \leq y \leq 1$ and 0 otherwise.

So I should say that:

**P(X^2>=1-y) = 1-P(X^2<=1-y) = 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) = 1 - [P(X>= - sqrt(1-y)) + P(X<=sqrt(1-y))] = 1-[ 1- P(X<= - sqrt( 1 -y) + P (X<= sqrt (1-y))]**

=P(X<= -sqrt(1-y)) - P(X<= sqrt(1-y))

Re: Probability Density Function Question

But to find P(X<= - sqrt(1-y)) and P(X<=sqrt(1-y)) I have to find the Cdf F X (x)

The thing that worries me is that since -1<=x<=1 if I integrate f X (x) from -1 to 1 I will get a real number and not a cdf in terms of x. So what should I do?

Then the cdf will be f Y (y) = d (F Y (y)) / dy and is a matter of calculation.

Re: Probability Density Function Question

Quote:

Originally Posted by

**Darkprince** But to find P(X<= - sqrt(1-y)) and P(X<=sqrt(1-y)) I have to find the Cdf F X (x)

The thing that worries me is that since -1<=x<=1 if I integrate f X (x) from -1 to 1 I will get a real number and not a cdf in terms of x. So what should I do?

Then the cdf will be f Y (y) = d (F Y (y)) / dy and is a matter of calculation.

The cdf is given by $\displaystyle \int_{-1}^{-\sqrt{1-y}} \frac{3}{8} (1 - x)^2 dx + \int^{1}_{\sqrt{1-y}} \frac{3}{8} (1 - x)^2 dx$.

Your job is to do the integrations and then differentiate the result with respect to y.

Re: Probability Density Function Question

Quote:

Originally Posted by

**mr fantastic** The cdf is given by $\displaystyle \int_{-1}^{-\sqrt{1-y}} \frac{3}{8} (1 - x)^2 dx + \int^{1}_{\sqrt{1-y}} \frac{3}{8} (1 - x)^2 dx$.

Your job is to do the integrations and then differentiate the result with respect to y.

If I say that I have to do 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) could I do (1 - the integral from -sqrt(1-y) to +sqrt(1-y) of the cdf)? i.e just one integral?

Re: Probability Density Function Question

Quote:

Originally Posted by

**Darkprince** If I say that I have to do 1- P(-sqrt(1-y) <=X<= +sqrt(1-y)) could I do (1 - the integral from -sqrt(1-y) to +sqrt(1-y) of the cdf)? i.e just one integral?

Yes.

Re: Probability Density Function Question