In this video, Blitzstein says that some random variables have no pdf but do have a cdf. Also, in my course material, I studied that converging in mean was stronger than converging in cdf which itself was stronger than converging in pdf.

I understand the maths behind it but I cannot think of an intuitive explanation for this.

  • $\begingroup$ The distribution illustrated at stats.stackexchange.com/a/104018/919 is a little unusual -- and for that reason helps inform one's intuition. It has a CDF (which is shown); it assigns positive probability to all open intervals between $0$ and $1;$ yet it has no PDF. $\endgroup$
    – whuber
    Commented Nov 24, 2020 at 22:06
  • 2
    $\begingroup$ I must say "converging in cdf" and "converging in pdf" are not standard terminologies. Can you clarify their exact mathematical meanings? $\endgroup$
    – Zhanxiong
    Commented Nov 16, 2023 at 4:56

1 Answer 1


I took a look at the link. Here's my interpretation:

Any random variable $X$ has a CDF, whether discrete or continuous, as the CDF is simply the probability that your random variable takes on a value less than or equal to some fixed value, $x$. Specifically, the CDF of $X$, $F_X$ is defined as $F_X(x) = P(X \leq x)$.

However, $X$ must be continuous to have a PDF. Of course, one could say that a discrete random variable has a PDF, too. It just has "infinite density" at all points with some probability mass and zero elsewhere! But, it's more useful to think of a discrete random variable as having a probability mass function instead. I think this is where the professor makes a distinction.

To address your second question, take the canonical example of a sequence of random variables $Y_n$ that takes on the value $n$ with probability $1/n$ and $0$ with probability $1-(1/n)$. You can verify that $Y_n \rightarrow 0$ in probability (and therefore converges in distribution too), yet $E(Y_n) = 1$ always.

Converging in probability (and, by extension, convergence in distribution) is weaker than convergence in mean precisely because of this previous example! Intuitively, convergence in distribution and probability simply rely on probabilities; in the previous example, all that mattered is that the "amount" of probability mass that was "away from zero" was shrinking to zero. Convergence in mean also relies on "where" that stray probability mass is, as well. As per my example, you can keep throwing your probability mass further and further from zero to counteract the fact that the probability of $Y_n > 0$ was shrinking. It's because convergence in mean controls "where the mass is" that makes it a stronger mode of convergence.

Finally, convergence in distribution is actually weaker than convergence in density/mass functions. Take the example of

$$F_{X_n}(x) = x - \frac{\sin(2\pi n x)}{2\pi n x}$$

and its associated density on (0,1), say. Clearly $X_n \rightarrow X$, where $X \sim \mathcal{U}(0,1)$ in distribution. Differentiating, we quickly find that the density doesn't even have a pointwise limit!

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    $\begingroup$ Thank you for this very clear and well written answer $\endgroup$ Commented Nov 25, 2020 at 10:04

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