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I would like to isolate p in the following. I am not sure if it is even possible.

a = B(1; 10, p)

B(x;n,p) is the cumulative binomial function.

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Is this the cumulative distribution? If so you should be able to figure out p (numerically at least, you are inverting a degree 10 polynomial in $p$). –  copper.hat May 18 '12 at 6:58
    
Sorry, I am really not good at math at all. To be honest, I was hoping for a formula. Indeed I don't understand your hint. –  Norham Vikser May 18 '12 at 7:15
    
If I understand you correctly, then you have (expanding the formula) $a = (1-p)^{10}+10 p (1-p)^9$. For a given $a$ and for this particular expression, there may be 0, 1 or 2 solutions (ie, values of p that satisfy the expression). –  copper.hat May 18 '12 at 7:23

1 Answer 1

up vote 2 down vote accepted

The aim is to solve $a=u(p)$ with $u(p)=(1-p)^{10}+10p(1-p)^9=(1+9p)(1-p)^9$. Since $u$ is a high degree polynomial, there can exist no formula inverting it in full generality and using only usual functions. However...

Since $u'(p)\lt0$ for every $p$ in $(0,1)$, the function $u$ is decreasing on $[0,1]$ from $u(0)=1$ to $u(1)=0$. Hence, for each $a$ in $(0,1)$, there exists a unique value $p_a$ in $(0,1)$ such that $u(p_a)=a$.

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I think $u$ has a maximum in $[0.7,0.8]$? $u'(0.7)>0$, $u'(0.8) <0$. –  copper.hat May 18 '12 at 7:40
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@copper.hat: Well... no. –  Did May 18 '12 at 7:47
    
@copper.hat pretty clear from the derivative: $\frac{d}{dp}((1-p)^{10}+10 p (1-p)^9) = -90 (p-1)^8 p$ –  Drew Christianson May 18 '12 at 7:56
    
My apologies, I had $u(p) = (1+9p)(1-p^9)$. (Which explains my flipperoo above. I thought it was invertible and then messed up and changed my mind.) –  copper.hat May 18 '12 at 8:02
    
I have realized that there exists a unique value fulfilling this. I also got to the (1+9p)(1-p)^9. I need the function for a script, so I was hoping that a "p = u_inv(a)" could be constructed. But I guess not. –  Norham Vikser May 18 '12 at 9:31

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