How to Find a Probability Density Function (PDF)

A probability density function (PDF) describes an absolutely continuous distribution of a random variable. The method for finding one depends on what information you have: a cumulative distribution function (CDF), a formula for the random variable, or a named distribution. Without some such information, there is no unique PDF to find.

What a probability density function tells you

For a random variable $X$ with a PDF $f_X(x)$, the density is nonnegative and gives probabilities by integration:

$$P(a \leq X \leq b)=\int_a^b f_X(x)\,dx.$$

The total probability must be one:

$$\int_{-\infty}^{\infty} f_X(x)\,dx=1.$$

A density value is not itself a probability. In particular, for a variable with a PDF, $P(X=x)=0$ for any single value $x$, even if $f_X(x)$ is positive. Probabilities come from areas under the density curve.

Method 1: Differentiate the CDF

If you know the CDF $F_X(x)=P(X\leq x)$ and the distribution is absolutely continuous (so it has a PDF), then its PDF is the derivative of the CDF almost everywhere:

$$f_X(x)=\frac{d}{dx}F_X(x).$$

First identify the intervals on which the CDF changes, differentiate on those intervals, and set the density to zero outside the range of possible values. A CDF can have corners where its derivative is undefined; that does not prevent the random variable from having a density. For an absolutely continuous distribution, the derivative formula applies almost everywhere. But differentiating a CDF does not recover discrete or singular components of a distribution.

Worked example: the square of a uniform variable

Suppose $X$ is uniform on $[0,1]$, and define $Y=X^2$. Since $0\leq X\leq1$, the possible values of $Y$ are also in $[0,1]$.

For $0\leq y\leq1$,

$$\begin{aligned} F_Y(y)&=P(Y\leq y)\\ &=P(X^2\leq y)\\ &=P(X\leq\sqrt{y})\\ &=\sqrt{y}. \end{aligned}$$

Including the parts outside the support, the CDF is $0$ for $y<0$, $\sqrt{y}$ for $0\leq y\leq1$, and $1$ for $y>1$. Differentiate on the interior of the support:

$$f_Y(y)=\begin{cases} \dfrac{1}{2\sqrt{y}},&0<y<1,\\ 0,&\text{otherwise.} \end{cases}$$

The density becomes unbounded as $y$ approaches zero, but it is still valid: the singularity is integrable, and $\int_0^1 \frac{1}{2\sqrt{y}}\,dy=1$. The value assigned at a single endpoint does not change any probabilities.

Method 2: Transform a known density

If $Y=g(X)$ and $g$ is one-to-one and differentiable on the support of $X$, use the change-of-variables formula. Write $x=g^{-1}(y)$, then

$$f_Y(y)=f_X\bigl(g^{-1}(y)\bigr)\left|\frac{d}{dy}g^{-1}(y)\right|,$$

for values of $y$ in the transformed support; set it to zero elsewhere. The absolute value accounts for whether the transformation increases or decreases.

If $g$ is not one-to-one, include every solution $x_i$ to $g(x_i)=y$ that lies in the support of $X$:

$$f_Y(y)=\sum_i \frac{f_X(x_i)}{|g'(x_i)|}.$$

This formula assumes the relevant branches are differentiable and that the derivative is nonzero at the inverse points. When a transformation has endpoints or stationary points where this fails, derive the CDF directly or handle those points separately.

For the worked example, the inverse on $[0,1]$ is $x=\sqrt{y}$, and $dx/dy=1/(2\sqrt{y})$. Since $f_X(x)=1$ on $[0,1]$, the transformation formula gives the same density, $f_Y(y)=1/(2\sqrt{y})$ for $0<y<1$.

Method 3: Recognize a named distribution

Sometimes the PDF is already known once you identify the model and its parameters. For example, a normal random variable with mean $\mu$ and standard deviation $\sigma>0$ has density

$$f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right),\qquad -\infty<x<\infty.$$

Check the parameter convention: some distributions are written using variance rather than standard deviation, or use a rate parameter where another source uses a scale parameter. If the random variable is a sum or another function of random variables, you may need a convolution, a transformation, or a CDF calculation rather than a standard named formula.

Common mistakes

• Treating a density value as a probability. To find an interval probability, integrate the density over that interval.
• Forgetting the support. A formula may only apply on a specific interval; the density is zero outside it unless the model says otherwise.
• Omitting the derivative factor. When transforming a variable, the inverse derivative (or its absolute value) is essential.
• Using only one inverse branch. For a many-to-one transformation such as $Y=X^2$ when $X$ can be positive or negative, contributions from both roots must be included.
• Assuming every random variable has a PDF. Discrete variables have probability mass functions. Some distributions combine point masses with a continuous density, and singular continuous distributions also exist, so their entire probability cannot be represented by an ordinary PDF alone.

After finding a candidate PDF, check that it is nonnegative on its support and that its integral over the full support is one. If either check fails, revisit the differentiation, transformation, or support.