How to find the following density function? Suppose $X$ is distributed exponentially with parameter $\lambda$. 
Define $Y=X^2$.
How would I find out $f_Y(t)$ the density function of $Y$.
Any help?
 A: $F_Y(y) = P\{Y \leq y\} = P\{X^2\leq y\} = P\{-\sqrt{y} \leq X \leq \sqrt{y}\} = F_X(\sqrt{y})- F_X(-\sqrt{y})$. Can you take it from here?
A: Here $X$ is an exponential distribution random variable with parameter $\lambda$,
$$f_X (x) = \lambda e^{-\lambda x}, $$ 
Then by change of variable, we have:
$$Y = g(X) = X^2$$
$$x = g^{-1}(y) = y^{\frac{1}{2}},$$
$$ \frac{d[g^{-1}(y)]}{dy} = \frac{1}{2} y^{-\frac{1}{2}}$$
and by formula (from change of variable technique): 
$$f_Y (y) = f_X (g^{-1}(y)) |\frac{d[g^{-1}(y)]}{dy}|,$$
$$ \implies f_Y (y) =\lambda e^{-\lambda y^{\frac{1}{2}}} \frac{1}{2} y^{-\frac{1}{2}} = \frac{\lambda}{{2} {\sqrt{y}}} e^{-\lambda \sqrt{y}}.$$
Alternative approach:
The cumulative distribution function (CDF) is 
$$F_Y(y) = P\{Y=X^2 \leq y\} = P\{-\sqrt{y} \leq X \leq \sqrt{y}\} = F_X(\sqrt{y}) \\ = \int_{0}^{\sqrt{y}} f_X(x)dx = \int_{0}^{\sqrt{y}} \lambda e^{-\lambda x} dx = 1 - e^{- \lambda \sqrt{y}},$$
then for probability density function (PDF) from the derivative of $F_Y(y)$ w.r.t $y$ we have 
$$\implies f_Y(y) = \frac{d[F_Y(y)]}{dy} = \frac{\lambda}{{2} {\sqrt{y}}} e^{-\lambda \sqrt{y}}.$$

Note: Change of variable technique
Let $X$ be a continuous random variable with probability density function $f_X(x)$ over an interval $S$. Suppose the mapping $Y = g(X)$ where $g: S \rightarrow T$ is a differentiable function. 


*

*If $g$ is a strictly increasing function:
$$F_Y (y) = \mathbb{P}(Y \leq y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \leq g^{-1}(y)) = F_X [g^{-1}(y)]$$
such that $y \in T$. Then, by taking derivatives w.r.t $y$ we have 
$$f_Y(y) = \frac{d[F_Y(y)]}{dy} = f_X{(g^{-1}(y))}\frac{d[g^{-1}(y)]}{dy} $$

*If $g$ is a strictly decreasing function:
$$F_Y (y) = \mathbb{P}(Y \leq y) = \mathbb{P}(g(X) \leq y) = \mathbb{P}(X \geq g^{-1}(y)) = 1 - F_X [g^{-1}(y)]$$
such that $y \in T$. Then, by taking derivatives w.r.t $y$ we have 
$$f_Y(y) = \frac{d[F_Y(y)]}{dy} = -f_X{(g^{-1}(y))}\frac{d[g^{-1}(y)]}{dy}.$$
A: In the more general case: Let $Y = g(X)$ and let 
$$f_X(x) = \frac{\mathrm{d}}{\mathrm{d}x}\mathbb{P}(X < x)$$
assuming it exists. 
Then,
$$\mathbb{P}(Y  < y) = \mathbb{P}( g(X)  < y) = \mathbb{P}( X \in g^{-1}((-\infty,y)) )$$
Now let $ g^{-1}((-\infty,y)) = (a_1,b_1) \cup(a_2,b_2) \cup (a_3,b_3) \cup \ldots  $, where $a_1<b_1<a_2<b_2<a_3<b_3<\ldots$
Then,
\begin{align}
\mathbb{P}( Y < y)
&= \mathbb{P}( X \in (a_1,b_1) ) + \mathbb{P}( X \in (a_1,b_2) ) +\mathbb{P}( X \in (a_3,b_3) ) + \ldots\\
&= \mathbb{P}( X < b_1) - \mathbb{P}( X < a_1) 
+\mathbb{P}( X < b_2) - \mathbb{P}( X < a_2)
+\mathbb{P}( X < b_3) - \mathbb{P}( X < a_3)+ \ldots\\
\Rightarrow 
f_Y(y) = \frac{\mathrm{d}}{\mathrm{d}y}\mathbb{P}(Y< y)
 &= \left(\frac{\mathrm{d}b_1}{\mathrm{d}y}\right) \frac{\mathrm{d}}{\mathrm{d}b_1} \mathbb{P}( X < b_1)-\left(\frac{\mathrm{d}a_1}{\mathrm{d}y}\right) \frac{\mathrm{d}}{\mathrm{d}a_1} \mathbb{P}( X < a_1)  + \ldots\\
&= \left(\frac{\mathrm{d}b_1}{\mathrm{d}y}\right) f_X(b_1)-\left(\frac{\mathrm{d}a_1}{\mathrm{d}y}\right) f_X(a_1)  + \ldots
\end{align}
In your case, $f_X(x) = \lambda e^{-\lambda x}\mathbb{1}_{x > 0}$, and $g(x) = x^2 \Rightarrow g^{-1}((-\infty, y)) = (-\sqrt{y},\sqrt{y})$. So 
$$f_Y(y)= \left(\frac{\mathrm{d}\sqrt{y}}{\mathrm{d}y}\right) \lambda e^{-\lambda \sqrt{y}}\mathbb{1}_{\sqrt{y} >0}-\left(\frac{\mathrm{d}(-\sqrt{y})}{\mathrm{d}y}\right) \lambda e^{\lambda\sqrt{y}}\mathbb{1}_{-\sqrt{y} >0} = \frac{\lambda}{2\sqrt{y}} e^{-\lambda \sqrt{y}}$$
