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- Jan 26, 2012

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This is something that when I see the work done it makes sense, but I find it difficult to do myself. I'm also aware there is an explicit formula for doing this but that involves Jacobians and a well-defined inverse, so I think it's more intuitive to do it step-by-step.

$P[Y \le y]=P[X^2 \le y]=P[X \le \sqrt{y}]$

Is this the direction I should be heading? From here should I try differentiating?

EDIT: I found some material on this question. It's important to note that $Y$ cannot be less than 0 so it seems that:

$$P[X \le \sqrt{y}]=P[-\sqrt{y} \le X \le \sqrt{y}]$$

Correct? Do the square roots look strange to anyone else? On my computer they look horrible.

**Problem**: Suppose $X \text{ ~ } U(0,1)$. Find the density function of $Y=X^2$.**My attempt**: I believe we should start by transforming the CDF and then differentiate to find the density function.$P[Y \le y]=P[X^2 \le y]=P[X \le \sqrt{y}]$

Is this the direction I should be heading? From here should I try differentiating?

EDIT: I found some material on this question. It's important to note that $Y$ cannot be less than 0 so it seems that:

$$P[X \le \sqrt{y}]=P[-\sqrt{y} \le X \le \sqrt{y}]$$

Correct? Do the square roots look strange to anyone else? On my computer they look horrible.

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