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I would like to double check my approach to this problem: Let $X$ be uniform $\sim [0,1]$. For the random variable $Y=\sqrt X$, find the PDF and CDF. I got the CDF to be: $$ F_Y(x) = \begin{cases}0 ,& \text{if }\; x ≤ 0,\\ x^2,& \text{if }\; 0 ≤ x≤ 1,\\ 1, &\text{if }\; x ≥ 1. \end{cases}$$ To find the PDF, differentiate the CDF, right? But I am a bit lost on how to do this. I have scoured my textbook for hours and would appreciate some sort of direction. Thank you!

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  • $\begingroup$ It's a piecewise function. You differentiate each piece to get $f'(x)$ over that piece. $\endgroup$ Commented Feb 6, 2023 at 0:19
  • $\begingroup$ @Semiclassical So the PDF would be 2x? Also to clarify - the CDF is presented as piecewise but the PDF is not, right? Sorry if this is a really dumb question. I am just struggling to grasp the obvious and my teacher literally did not take us through any examples $\endgroup$
    – user1147005
    Commented Feb 6, 2023 at 0:24
  • $\begingroup$ For $0\leq x\leq 1$, yes. (Though to avoid ambiguity, I'd prefer to let $Y=\sqrt{X}$ and therefore $P(Y\leq y)=y^2\implies dP/dy = 2y$. It's the same function but emphasizes the transformation.) $\endgroup$ Commented Feb 6, 2023 at 0:26
  • $\begingroup$ @Semiclassical For the other parts, the PDF is just zero I believe? Would I need to specify that on the solution? Sorry for all the questions, just want to clarify how a solution to this type of question should be presented. $\endgroup$
    – user1147005
    Commented Feb 6, 2023 at 0:29

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You just differentiate the CDF; since it is a piece-wise function, you differentiate it separately at each interval. The result will be

$$ f_Y(x)=\frac{d}{dx} F_Y(x)=\begin{cases} 0 &\text{if } x<0;\\ 2x &\text{if } 0\le x\le 1;\\ 0 &\text{if } x>1 \end{cases} =\begin{cases} 2x &\text{if } 0\le x\le 1;\\ 0 &\text{otherwise}. \end{cases} $$

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    $\begingroup$ Thank you for verifying, this is what I thought! $\endgroup$
    – user1147005
    Commented Feb 6, 2023 at 16:28

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