To show the desired result, we are working with a continuous random variable $X$ with probability density function (p.d.f.) $f_X(x)$, and we define a new random variable $Y$ as a strictly monotone and differentiable function of $X$, i.e.,

$Y = g(X),$

where $g$ is a strictly monotone differentiable function.

Our goal is to find the p.d.f. of $Y$, denoted as $f_Y(y)$.

Step 1: Change of Variables

We begin by recalling the general formula for the probability density function transformation under a monotone function. If $Y = g(X)$, then for a continuous random variable, the p.d.f. of $Y$, $f_Y(y)$, is related to the p.d.f. of $X$, $f_X(x)$, by the formula:

$f_Y(y) = f_X(x) \left| \frac{dx}{dy} \right|.$

The absolute value is taken because the derivative could be negative if the function is decreasing.

Step 2: Inverse Relationship

To apply this formula, we need to express $x$ as a function of $y$. Since $g$ is strictly monotone (so it is invertible), we can write $x = g^{-1}(y)$.

Step 3: Differentiation of the Inverse

Next, we calculate the derivative of $x$ with respect to $y$. Using the chain rule, we have:

$\frac{dx}{dy} = \frac{d}{dy} g^{-1}(y) = \frac{1}{g'(x)},$

where $g'(x)$ is the derivative of the function $g(x)$.

Step 4: Substituting into the Formula

Now, using the inverse relationship $x = g^{-1}(y)$ and the derivative formula, we substitute into the formula for $f_Y(y)$:

$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{1}{g'(g^{-1}(y))} \right|.$

Since $g'(x)$ is the derivative of the function $g(x)$, we can simplify the expression as:

$f_Y(y) = f_X(g^{-1}(y)) \frac{1}{|g'(g^{-1}(y))|}.$

Step 5: Final Expression

Thus, the probability density function of $Y$ is given by:

\begin{cases} f_X(g^{-1}(y)) \left| \frac{1}{g'(g^{-1}(y))} \right|, & \text{if } y = g(x) \text{ for some } x, \\ 0, & \text{otherwise}. \end{cases}$$ This completes the proof. ### Summary Given that $$Y = g(X)$$, where $$g$$ is a strictly monotone differentiable function, the p.d.f. of $$Y$$ is: $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{1}{g'(g^{-1}(y))} \right|.$$ If you need further clarification or have additional questions, feel free to ask. --- Here are some related questions that expand on this concept: 1. What happens to the p.d.f. of $$Y$$ if $$g(x)$$ is a decreasing function instead of increasing? 2. How would the formula for the p.d.f. change if $$g(x)$$ is not invertible? 3. What is the relationship between cumulative distribution functions (C.D.F.s) and p.d.f.s under transformation? 4. How would you apply this result to a specific example, such as $$Y = X^2$$ for a normally distributed $$X$$? 5. How would you derive the p.d.f. for a transformed random variable using C.D.F. methods instead? Tip: When transforming random variables, always check whether the function is invertible and whether you need to take absolute values of derivatives (especially for decreasing functions).