Find the PDF in terms of another variable

[CivilSigma]

Homework Statement

For

$$f_x(x)=4x^3 ; 0 \leq x \leq 1$$

Find the PDF for $$ Y < y=x^2$$

The Attempt at a Solution

So, we take the domain on x to be:

$$0\leq x \leq \sqrt y$$

and integrate:

$$ \int_0^{\sqrt y} f_x(x) dx = \int_0^{\sqrt y} 4x^3 dx$$

Do we integrate with respect to x or y? I am not sure about this part of the problem.

If we solve the integral above we get:

$$f_x(y)=y^2 ; y \leq x^2 $$

Is this approach correct?

Thank you.

 

[RPinPA]

"So, we take the domain on $x$ to be:"

Wait, what? The domain of $x$ is already defined, it's (0,1]. Why would you change it?

Your integral will is calculating the cumulative distribution of $x$, the probability that $x$ is less than or equal to $\sqrt y$. That isn't what was asked for. However $P(X \leq \sqrt y)$ = $P(X^2 \leq y)$ = $P(Y \leq y)$. So you've calculated the cumulative distribution of $y$. IOnce you have the CDF for $y$ you can get the $PDF$.

Now there's a lot wrong with your concluding statement $f_x(y) = y^2; y \leq x^2$.
- It's not $f_x$ as this is supposed to be the density of $y$.
- As explained already, you didn't calculate a density, you calculated a CDF.
- Since $y = x^2$ and the domain of $x$ is (0, 1], the domain of $y$ is (0, 1]. The density of $y$ is not related to a particular value of $x$.

Working with the CDF is a useful thing to do, you're just misinterpreting what every single thing in this calculation is for.

Let's say we want the CDF of $y$, the probability $P(Y \leq y)$ for any value of $y$, which as we said could be anything in (0, 1] based on the domain of $x$. $Y \leq y \iff X^2 \leq y \iff X \leq \sqrt y$. With a different definition of $X$ that last event might be ($0 \leq X \leq \sqrt y$ or $-\sqrt y \leq X \leq 0$) but X can't be negative here.

OK, so we've established that $P(Y \leq y) = P(X \leq \sqrt y) = \int_0^{\sqrt y} 4x^3 dx$, and that should address your questions as to whether you're integrating with respect to $x$ or $y$, and also what you've calculated here and how it relates to what you were asked for.

To summarize: You decide to calculate the cumulative distribution of Y. You want to know how many Y's fall into a given range $(0,y]$, for any choice of $y$. You know what X values will give you Y's in the desired range, so you know how to use the density of X to calculate that probability. And that gives you an expression for the CDF of Y.

 

[CivilSigma]

This makes more sense.

So, we first find cumulative probability in terms of the constraint, and then differentiate to get the probability function.

So:

$$F_y(y) = y^2 ; 0 \leq y \leq 1$$
$$f_y (y) = \frac {F_y(y)}{dy} = 2y ; 0 \leq y \leq 1$$