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Given is $U \sim R(0,1)$ and $X=g(U) = -\frac{1}{\lambda} \ln(U), \space \space \lambda > 0$

What is the density of random variable $\sqrt{X}$?

($X \sim R(a,b)$ in general means equal distribution.)

Original question is from here Calculate the densitiy of the random variable $X$

But I wondered how you can do it correct and easy when there is little change, like here from $X$ to $\sqrt{X}$.

$$P(\sqrt{X} \leq x) = P\left(\sqrt{-\frac{1}{\lambda} \cdot \ln(U)} \leq x\right) = P\left(-\frac{1}{\lambda} \cdot \ln(U) \leq x^2\right) = P\left(\ln(U) \geq -\lambda x^2\right) = P\left(U \geq e^{-\lambda x^2}\right)$$

$$F_X(x) = \int_{e^{-\lambda x^2}}^{1} dU = 1-e^{-\lambda x^2}$$

$$f(x) = \frac{d}{dx}\left(1-e^{-\lambda x^2}\right) = -\lambda 2xe^{-\lambda x^2}$$

Is it correct like this?

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    $\begingroup$ Yes, except you made a mistake on the sign. So $f(x) = 2\lambda x e^{-\lambda x^2}$ $\endgroup$ – induction601 Dec 6 '17 at 21:08

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