MHB PDF of X^2: General Method & Question Answered

[nacho-man]

I just started this topic and have a question:

For a positive continuous random variable X, write down the PDF of Y = X^2 in terms of the PDF of X.

So; writing the PDF of X, I get

P(0<X<∞) =integral from 0 to ∞ fx(x)dx

Is this correct? For Y=X^2, wouldn't the pdf be the exact same thing, since the question asks for positive continuous random variables anyway?

Is there a general method, approach I should have about this. I find pdfs to be confusing, the textbooks are heavy on jargon.

 

[I like Serena]

I just started this topic and have a question:

For a positive continuous random variable X, write down the PDF of Y = X^2 in terms of the PDF of X.

So; writing the PDF of X, I get

P(0<X<∞) =integral from 0 to ∞ fx(x)dx

Is this correct? For Y=X^2, wouldn't the pdf be the exact same thing, since the question asks for positive continuous random variables anyway?

Is there a general method, approach I should have about this. I find pdfs to be confusing, the textbooks are heavy on jargon.

Welcome to MHB, nacho! :)

Since your questions were sufficiently different to warrant their own threads, I have taken the liberty to split them into 2 threads. We prefer 1 question per thread here (unless the questions are parts of the same problem statement).
Now to address your problem, to find a PDF, the general method is to look at the CDF and take its derivative.

In your case the CDF of Y is:
$$P(Y < y) = P(X^2 < y) = P(X < \sqrt y) = \int_{0}^{\sqrt y} f_X(x) dx$$
Note that the lower boundary can be zero, since X is given to be positive.

To find the PDF of Y, you need to differentiate its CDF.
$$f_Y(y) = \frac{d}{dy}P(Y < y)$$
Now supposed $F_X(x)$ is the anti-derivative of $f_X(x)$, which is actually the same as the CDF of X, then you get:
$$f_Y(y) = \frac{d}{dy}P(Y < y) = \frac{d}{dy}\int_{0}^{\sqrt y} f_X(x) dx
=\frac{d}{dy}\Big(F_X(\sqrt y) - 0\Big) = f_X(\sqrt y) \frac{d}{dy}(\sqrt y)$$

Does that answer your question?