For a given density function, how does one find its distribution function?

For example, I have a density function:

$f(x)= \begin{cases} t ^2 / 9 & \text{if } t \in (0,3)\\ 0 & \text{otherwise} \end{cases}$

How do I find the probability distribution function?

Do I just need to find the derivative for every equation in it?


The cumulative distribution function (CDF) is the anti-derivative of your probability density function (PDF). So, you need to find the indefinite integral of your density. Only if you are given the CDF, you can take its first derivative in order to obtain the PDF. However, your proposed function is not a density function because a density has the property that it integrates to one. This does not hold here:

$\int_0^3 9t^2 dt = [3t^3]_0^3 = 3 \cdot 3^3 = 3^4 = 81.$

So, this is not a density function and you either need to adjust the boundaries or scale it by $\frac{1}{81}$. An adjustment of the boundaries could be achieved by changing the upper bound $3$ to $\sqrt[3]{\frac{1}{3}}$.

EDIT: The OP adjusted the formula after my remark above to a PDF

Now, your function is a density. See, it integrates to one:

$\int_0^3 \frac{1}{9}t^2 dt = [\frac{1}{27}t^3]_0^3 = \frac{1}{27}3^3 = \frac{1}{27}27= 1.$

For getting the CDF $F(t)$, you need to find the indefinite integral:

$F(t) = \int \frac{1}{9}t^2 dt = \begin{cases} \frac{1}{27}t^3 & \text{if } t \in (0,3)\\ 0 & \text{otherwise} \end{cases}$

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  • 2
    $\begingroup$ -1 The CDF is not the antiderivative or indefinite integral of the pdf; it is the definite integral -- the area under the pdf curve. Nor is it true that $F(t)$ equals $0$ when $t \geq 3$: $F(t)$ has value $1$ for all $t \geq 3$. For any real number $x$, $$F(x) = \int_{-\infty}^x f(t) \, \mathrm dt,$$ the area under the pdf curve to the left of $x$ (and including the point $x$ since the CDF $F(x)$ is defined as being the value of the probability that the random variable has value less than or equal to $x$ -- call it "no larger than $x$" if you like. $\endgroup$ – Dilip Sarwate Dec 13 '18 at 16:32

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