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I have this problem to determine a density function $f_{Y}(y)$ of $Y$ given $X$ as a random variable with uniform distribution on $[−1, 1]$ and $Y = X^2$. So: $f_X(x) = \frac{1}{2}$ for $x \in [-1, 1]$. I approached this problem with two different ways, 1) through determining first the $F_Y$ as the $CDF$, then derive from it $PDF$ 2) by transformation.

For the first method I got $f_Y(y) = \frac{1}{2\sqrt(y)}$ for $y \in ]0, 1]$ This seems correct. However, for transformation method, I tried the following:

$g(x) = x^2$ and $g'(x) = 2x$

$x_1 = g^{-1}(y) = \sqrt(y)$

so $f_Y(y) = \frac{f_X(x_1)}{|g'(x_1)|} = \frac{1/2}{2\sqrt(y)} = \frac{1}{4\sqrt(y)}$

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Let $Z=X^2$, $z \geq 0$, $$P(Z \leq z) = P(X^2 \leq z) = P(X \leq \sqrt{z}, -X \leq \sqrt{z})$$ $$= P(X \leq \sqrt{z}, X \geq -\sqrt{z}) = F_X(\sqrt{z})-F_X(-\sqrt{z})$$

$$f_Z(z) = \frac{\partial P(Z \leq z)}{\partial z} = \frac{f_X(\sqrt{z})+f_X(-\sqrt{z})}{2\sqrt{z}}$$

If you derive the density of $Z$ directly using density of $X$, then u will get the same answer where $\frac{1}{2\sqrt{z}}$ is the jacobian. For $Z = g(X)$ with $g^{-1}$ continuously differentiable and $g$ is continuous, the density: $$f_Z(z) dz = P(Z \in [z,z+dz]) = P(X \in g^{-1}([z,z+dz]))$$ $$= \sum f_X(g^{-1}(z)) \times length(g^{-1}([z,z+dz]) = \sum_{x_i \in g^{-1}(z)} f_X(x_i) |\frac{\partial{g_i^{-1}(z)}}{\partial z}| dz$$ Where $g_i^{-1} = g|_{S_i}^{-1}, S_i = [x_i,x_i+\epsilon)$. The last step in the above derivation uses the mean value theorem and continuous differentiability of $g^{-1}(z)$. Now directly deriving the density, using above formula: $$f_Z(z) = (f_X(\sqrt{z})+f_X(-\sqrt{z})) |\frac{\partial \sqrt{z}}{\partial z}| = (f_X(\sqrt{z})+f_X(-\sqrt{z})) |\frac{1}{2\sqrt{z}}|$$.

So both gives the same answer.

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  • $\begingroup$ Thank you, but your explanation is not clear to me, I modified my question. $\endgroup$
    – Papa
    Dec 8, 2022 at 5:06
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    $\begingroup$ $dz$ is considered a very small increment, $P(Z \in [z,z+dz])$ means probability $Z$ is in the small interval $[z,z+dz]$. Since the interval is small the density function $f_Z(z)$ can be considered to be a constant function in this interval and we approximate it by $f_Z(z) dz = P(Z\in [z,z+dz])$ where $f_Z(z) dz$ is the integration of density $f_Z(z)$ over the interval $[z,z+dz]$ which is by definition $P(Z \in [z,z+dz])$. This approximation becomes exact when at the end you let $dz \rightarrow 0$. $\endgroup$
    – Balaji sb
    Dec 9, 2022 at 3:40
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    $\begingroup$ Note that in the previous comment we assumed $f_Z(z)$ to be constant in the interval $[z,z+dz]$ and integrated it over this interval to get $f_Z(z) dz$ as probability $Z$ is in $[z,z+dz]$. So similarly for finding $P(X \in g^{-1}([z,z+dz]))$, we do the same and assume $f_X(x)$ to be constant in the interval $g^{-1}([z,z+dz])$ and integrate it to get $f_X(g^{-1}(z)) \times length(g^{-1}([z,z+dz]))$. So Now we have $P(X \in g^{-1}([z,z+dz])) = f_X(g^{-1}(z)) \times length(g^{-1}([z,z+dz]))$. $\endgroup$
    – Balaji sb
    Dec 9, 2022 at 3:47
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    $\begingroup$ Now derivative can be thought of as slope and now we finllay do the approximation $length(g^{-1}([z,z+dz])) = |\frac{\partial g^{-1}(z)}{\partial z}| dz$ (since derivative is the slope, this works). $\endgroup$
    – Balaji sb
    Dec 9, 2022 at 3:51
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    $\begingroup$ All these approximations becomes exact when we divide by $dz$ and let $dz \rightarrow 0$ due to the assumptions on $g,g^{-1}$ i.e., $g^{-1}$ is continuously differentiable and $g$ is continuous. $\endgroup$
    – Balaji sb
    Dec 9, 2022 at 3:53

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