I'll be giving some classes on probability theory later this year, and so I've been going through the textbook to check that I'm up to speed. I came across the following question:
The discrete random variable $X$ has the cumulative distribution function $\mathrm{F}$ defined by $$\mathrm{F}(x) = \left\{ \begin{array}{ccc} 0 & & x=0 \\ \frac{(x+k)^2}{16} & & x=1,2,3 \\ 1 & & x > 3 \end{array}\right.$$ Find the value of $k$.
As far as I can tell, there isn't enough information to solve the problem. Using the fact that $\mathrm{F}(x) = \mathrm{P}(X\le x)$, and so $\mathrm{P}(X=x)=\mathrm{F}(x)-\mathrm{F}(x-1)$, I have obtained $$\begin{eqnarray*} P(X=0) &=& 0 \\ \\ P(X=1) &=& \tfrac{1}{16}(1+k)^2 \\ \\ P(X=2) &=& \tfrac{1}{16}(3+2k) \\ \\ P(X=3) &=& \tfrac{1}{16}(5+2k) \\ \\ P(X=4) &=& \tfrac{1}{16}(1-k)(7+k) \\ \\ P(X \ge 5) &=& 0 \end{eqnarray*}$$
As far as I can tell, these satisfy $\mathrm{F}(x)=\mathrm{P}(X \le x)$ for all $x \in \mathbb{N}$, independently of $k$.
Since all probabilities must lies between $0$ and $1$, I imposed the conditions $0 \le \mathrm{P}(X=x) \le 1$ for all $x \in \mathbb{N}$. Solving these inequalities tells us that $-\frac{3}{2} \le k \le 1$.
The books says that $k=1$ is the answer, but I think that $k$ with $-\frac{3}{2} \le k \le 1$ are all valid answers.
What have I missed?!
I do not know whether the book defines the cdf properly, as being defined for all real $x$. If it does, then what we have is only a partial description of the cdf. However, the description is sufficient to determine $k$.
A cdf is continuous from the right, so from $F(x)=1$ for $x\gt 3$ we can conclude that $F(3)=1$. The formula for $F(3)$ then shows that $(3+k)^2=16$. That yields $k=1$ or $k=-7$. However, $k=-7$ is impossible. For if $k=-7$ then $F(1)\gt 1$.