Given a continuous random variable $x$ with CDF of $x^3$ for $0\le x\le 1$ (and $0$ for $x \lt 0$ and $1$ for $x \gt 1$, rank the median, mode and mean.

My attempt: To find the mean, I first found the PDF to be $3x^2$. I then took $\int_0^1 x(3x^2) \,dx = \frac{3}{4}$

For the median, I set the CDF of $x^3$ equal to $\frac{1}{2}$ which is $\left(\frac{1}{2}\right)^\frac{1}{3}$

Finally, for the mode, I think that this would be the highest value of the PDF of $3x^2$ on the interval of $0 \le x \le 1$ , which should be $1$ but I am unsure if this reasoning is correct.

I am struggling to understand the concept of mode for continuous random variables since the probability of any individual point is $0$. I am unsure of whether my reasoning which is mostly carried over from discrete variables is applicable.

  • $\begingroup$ Here the mode is just the PDF's maximum. $\endgroup$ Jun 25, 2017 at 20:33

1 Answer 1


A mode represents the same quantity in continuous distributions and discrete distributions: The element in a random variable's domain at which the pdf is maximized. Sure, for continuous distributions you have to fudge the end of that a bit to something like "at which the pdf is locally maximized," but it's the same principle. In other words, your reasoning is valid.

In this case, our domain is the closed interval $[0,1]$, so the pdf $ 3x^{2}$ takes on a maximal value at either a critical point or at the endpoints $0,1$. The only critical point is $0,$ and $3x^{2}_{x=0} = 0$. Since $3x^{2}>0$ at $x=1$, your answer is correct: The mode is $1$.

Regarding how to interpret the meaning behind a mode of real-valued random variable, I wouldn't analyze it too much yet. ;) The bizarre, seemingly paradoxical idea of a real-valued random variable having zero probability at any isolated point can be resolved.

But it takes some analysis and topology to really get comfortable with that idea. After some of those classes, you can gleefully look back and interpret the idea more clearly and intuitively.


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