Consider a random variable $X$ with mass function $p_{X}(x)$ given by:

$$P(X=k) =\frac{6k^2}{n(n+1)(2n+1)}, \quad k= 1, \dots, n$$ I want determine the cumulative distribution function of $X$

We know that

$$F_{X}(k)=P(X \leq k)= P\left(\bigcup_{k' \leq k} \{X=k' \} \right)=\sum_{k' \leq k} p_{X}(k')$$

In our case this looks something like

$$F_{X}(x)=P(X \leq k)=\sum_{k'=1}^{k} P(X=k')=\frac{6k'^2}{k'(k'+1)(2k'+1)}$$

But something I don't understand is if $k$ goes from one to $n$, how does this influence the final result. How can I conclude the answer? any suggestion will be welcome!

  • $\begingroup$ Why have you replaced the $n$s in the mass function with $k'$ in the denominator of your cumulative distribution function? The random variable is $k$, not $n$. $\endgroup$
    – jbowman
    Nov 20, 2023 at 1:04
  • 1
    $\begingroup$ $F_{X}(k)=P(X \leq k)=\sum\limits_{k'=1}^{k} P(X=k')=\sum\limits_{k'=1}^{k}\dfrac{6k'^2}{n(n+1)(2n+1)}$ $=\dfrac{6}{n(n+1)(2n+1)}\sum\limits_{k'=1}^{k} k'^2$ at least for integer $k$ between $1$ to $n$. The sum over $k'$ is easy $\endgroup$
    – Henry
    Nov 20, 2023 at 1:08
  • $\begingroup$ @Henry If I am honest, I have been thinking for a while about the last sum but I don't see where is the facility for the computation, any help? $\endgroup$
    – Wrloord
    Nov 20, 2023 at 1:59
  • $\begingroup$ What do you mean by "facility for the computation"? $\endgroup$
    – jbowman
    Nov 20, 2023 at 2:28
  • 2
    $\begingroup$ As a shortcut, $\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$, as you should be able to deduce from your first equation! $\endgroup$
    – jbowman
    Nov 20, 2023 at 2:30


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