4
$\begingroup$

If a problem says to find the distribution of an RV, can I take that to mean the CDF, or take it to mean the PDF, OR is it ambiguous?

Here is a concrete example:

"If $X_1$ and $X_2$ are independent exponential random variables with respective parameters $a$ and $b$, find the distribution of $Z = \frac{X_1}{X_2}.$

$\endgroup$
1
  • 7
    $\begingroup$ Intuitively speaking, the distribution of a random variable $X$ is a complete set of information needed to describe all the probability-related questions regarding $X$. So, any of CDF, PMF, or PDF (whenever appropriate) would be a valid form of answer. In general, a distribution of a random variable $X$ refers to the pushforward probability measure $$A\mapsto\mathbf{P}(X\in A).$$ When $X$ is real-valued, this is completely characterized by the CDF. If the distribution is discrete or continuous, then it is completely characterized by PMF or PDF. $\endgroup$
    – Sangchul Lee
    Nov 14, 2022 at 2:34

1 Answer 1

Reset to default
5
$\begingroup$

The theory of probability is taught on so many levels of mathematical rigour, generality and abstraction that it is impossible to tell for sure without asking the author of the problem or the person responsible for the course.

In the general setting, the distribution of a random variable is the induced measure, that is, the pushforward of the probability measure defined in an abstract measurable space.

Commonly (or even exclusively) the term "random variable" refers to objects that take values in $\mathbb{R}^n$ or $\mathbb{R}$, where Borel sigma-algebra is defined. Then the distribution is described by a CDF (cumulative distribution function).

If the CDF is absolutely continuous, then one could use a PDF (probability density function) to describe the distribution of the random variable.

$\endgroup$
1
  • 1
    $\begingroup$ @Alborz For a Beta distribution, I would not be surprised to see the distribution stated as a density function rather than its integral. For a Cantor distribution, there is no density but the CDF can be stated. Each case fully specifies the distribution. $\endgroup$
    – Henry
    Nov 14, 2022 at 8:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .