Question

Let X be the length of the initial run in a random ordering of n ones and m zeros. That is, if the first k values are the same (either all ones or all zeros), then X Ú k. Find E[X].

Short answer

The $E\left[X\right\}$of the problem if X be the length of the initial run in a random ordering of n ones and m zeros $E\left[L\right]=\left(\frac{n}{m+1}\right)+\left(\frac{m}{n+1}\right)$.

Step-by-step solution

Given information

X= length of the initial run in a random ordering of n ones and m zeros

k-value are same

Solution

The solution will be shown below,

$\mathrm{E}\left[\mathrm{L}\right]=\mathrm{E}[\mathrm{L}\mid \text{First vlaue is one}]\frac{\mathrm{n}}{(\mathrm{n}+\mathrm{m})}$

$+\mathrm{E}[\mathrm{L}\mid \text{First value is}0]\frac{\mathrm{m}}{(\mathrm{n}+\mathrm{m})}$

Now if first value $=1$,

Length of run will be position of first $0$when considering remaining $n+m-1$values, of which $n-1$are one's and $m$are zero's

$\Rightarrow E\left[L\right]=\left(\frac{n+m}{m+1}\right)\left(\frac{n}{n+m}\right)+\left(\frac{n+m}{n+1}\right)\left(\frac{m}{n+m}\right)$

$\Rightarrow E\left[L\right]=\left(\frac{n}{m+1}\right)+\left(\frac{m}{n+1}\right)$

Final answer

The $E\left[X\right]$of the given problem Will be $E\left[L\right]=\left(\frac{n}{m+1}\right)+\left(\frac{m}{n+1}\right)$.