This may be a very easy question but when a CDF is differentiated, it becomes a pdf:

$${partial over partial z} F_Z(z) = f_Z(z)$$

But let's say I have a CDF of $F_Z(sqrt{z})$. Is the pdf this:

$${partial over partial z} F_Z(sqrt{z}) = f_Z(sqrt{z})$$

or this:

$${partial over partial z} F_Z(sqrt{z}) = f_Z(sqrt{z}) cdot {1 over 2 sqrt{z}} $$

**Contents**hide

#### Best Answer

You cannot create a CDF simply by taking a known valid CDF $F_Z(z)$ and replacing its argument $z$ by $sqrt{z}$. Remember that a CDF is a real-valued function defined for all values of its argument, and when you write $F_Z(sqrt{z})$, you are not telling us what value your new alleged CDF has when the *argument* of your new CDF has value smaller than $0$. Now, if $Z$ is a *nonnegative* random variable, then the function $$G(z) = begin{cases} F_Z(sqrt{z}), & z geq 0,\0, & z < 0,end{cases}tag{1}$$ is indeed a valid CDF **of a different random variable** $Y$ (it happens to equal $Z^2$) and if $F_Z(z)$ happens to be an absolutely continuous function, then the derivative of $G(z)= F_Y(z)$ which is $$frac{mathrm d}{mathrm dz}G(z) = g(z) = begin{cases} f_Z(sqrt{z})frac{1}{2sqrt{z}}, & z geq 0,\0, & z < 0,end{cases}$$ is the pdf of this different random variable.

### Similar Posts:

- Solved – Distribution of the convolution of squared normal and chi-squared variables
- Solved – CDF Variable Transformation
- Solved – Partial Derivative of Joint Distribution Function interpretation
- Solved – Partial Derivative of Joint Distribution Function interpretation
- Solved – MLE of $theta$ in $N(theta, theta^2)$