Probability of subintervals being less than x

Question

Alright so I was shown the equation $$\sum_{v=1}^n (-1)^{v+1} {n\choose v}\left(1-v\frac{x}{t}\right)^{n-1}_{+}. $$

in a previous post, as a method of finding the probability that a randomly broken stick with $n$ subintervals would have a length less than or equal to $x$, ($x$ is the length we want, $t$ is the length of the whole segment). I keep trying it for different $n$ with the $t=1$ and $x=0.5$ but I keep getting answers greater than one. What am I doing wrong here. I'll give an example of what I am doing for $n=5$.

For $v=1$ I get $5(0.5)^0=5$

For $v=2$ I get $-10(0)=0$

For $v=3$ I get $10(-0.5)^2=2.5$

For $v=4$ I get $-5(-1)^3=5$

Already I can see that the sum I am going to get at the end is not a probability. What am I doing wrong?

Link: A Broken Stick with $n$ breaks

It's a formula, not an equation. An "equation" has an equal sign in it somewhere. — Thomas Andrews, Jun 26 '17 at 20:40
Also, I think $(a)_{+}$ usually means $0$ if $a$ negative and $a$ otherwise, although I'm not sure about here. — Thomas Andrews, Jun 26 '17 at 20:46
Finally, notice the exponent is $n-1$, not $v-1$, so the first term is $5\cdot (0.5)^4$. — Thomas Andrews, Jun 26 '17 at 20:47

score 1 · Accepted Answer · answered Jun 26 '17 at 20:57

You've been computing:

$$\sum_{v=1}^n (-1)^{v+1} {n\choose v}\left(1-v\frac{x}{t}\right)^{v-1}.$$

But the correct formula is:

$$\sum_{v=1}^n (-1)^{v+1} {n\choose v}\left(1-v\frac{x}{t}\right)^{\color{red}{n-1}}_{\color{red}+}.$$

The link you've given indicates exactly what $(x)_{+}$ means:

$$(x)_{+}=\begin{cases}x&x>0\\0&x\leq 0\end{cases}$$

So when $t=1,x=0.5$, $1-v\frac{x}{t}\leq 0$ for $v\geq 2$ and the above formula gives:

$$n\cdot(0.5)^{n-1}$$

Probability of subintervals being less than x

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