# Determine cumulative distribution function

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!

• 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 $n$. Nov 20, 2023 at 1:04
• $F_{X}(k)=P(X \leq k)=\sum\limits_{k'=1}^{k} P(X=k')=\sum\limits_{k'=1}^{k}\dfrac{6k'^2}{n(n+1)(2n+1)}$ $=\dfrac{6}{n(n+1)(2n+1)}\sum\limits_{k'=1}^{k} k'^2$ at least for integer $k$ between $1$ to $n$. The sum over $k'$ is easy Nov 20, 2023 at 1:08
• @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? Nov 20, 2023 at 1:59
• What do you mean by "facility for the computation"? Nov 20, 2023 at 2:28
• 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! Nov 20, 2023 at 2:30