Find the PDF in terms of another variable

In summary, to find the PDF for ##Y < y = x^2##, we first calculate the CDF by taking the integral of the density of ##X## with respect to the constraint, which gives us the CDF of ##Y##. Then, we differentiate the CDF to get the PDF, giving us the desired probability function.
  • #1
CivilSigma
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58

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.
 
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  • #2
"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.
 
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  • #3
This makes more sense.

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

So:

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

Related to Find the PDF in terms of another variable

1. What is a PDF in terms of another variable?

A PDF, or probability density function, is a mathematical function that describes the probability of a random variable taking on a certain value. It is often used to model continuous random variables, and can be expressed in terms of another variable, such as time or distance.

2. How is a PDF related to other statistical concepts?

A PDF is closely related to other statistical concepts such as the cumulative distribution function (CDF) and the probability mass function (PMF). The CDF is the integral of the PDF and represents the probability that a random variable is less than or equal to a certain value. The PMF is the discrete version of the PDF and is used to model discrete random variables.

3. What is the significance of finding the PDF in terms of another variable?

Finding the PDF in terms of another variable allows us to understand the relationship between two variables and how they affect each other. It also allows us to make predictions or calculate probabilities for one variable based on the other. This can be useful in various fields such as finance, engineering, and biology.

4. How do you find the PDF in terms of another variable?

The process of finding the PDF in terms of another variable depends on the specific problem and the type of random variable being modeled. In general, it involves using mathematical techniques such as integration and differentiation to manipulate the given information and derive the desired PDF.

5. Can the PDF in terms of another variable be used to make predictions?

Yes, the PDF in terms of another variable can be used to make predictions or calculate probabilities for the random variable of interest. By plugging in different values for the other variable, we can determine the probability of the random variable taking on those values. However, it is important to note that the accuracy of these predictions depends on the accuracy of the given information and the assumptions made in the model.

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