# Extrapolation

• Apr 9th 2013, 10:58 AM
StevenDR
Extrapolation
Hello,

Lets say i'm measuring the probability of something happening at different stages.

For x = 1 I know the probability +- 0% (exact)
For x = 2 x = 3 and x = 4 I know the probability +- 8%

How well would i be able to say what the probability is at x = 5? and x = 6? And even further? (maximum to x = 10)
Also, I expect the probability to decreases exponentially.

If someone could help me with this that would be really awesome. http://www.mymathforum.com/images/sm...on_biggrin.gif
• Apr 9th 2013, 11:49 AM
Re: Extrapolation
If the probability is decreasing exponentially, why is it constant at $x = 2,3,4$ ?
Assuming it starts strictly decreasing at $x = 4$, i.e. $P(x = 4) > P(x = 5)$, then:

Let $n = x - 4$, and $P_n = ar^n$ where $n \geq 0$ and $0 < r < 1$
Since $P(x = 4) = P_{n = 0} = 0.08$, it follows that $a = 0.08$
So $P_n = (0.08)r^n$

We also need $\sum_{x=1}^{\infty}{P(x)} = 1$ because we are dealing with probabilities.
$\sum_{x=1}^{\infty}{P(x)}$
$P(x = 1) + P(x = 2) + P(x = 3) + \sum_{n=0}^{\infty}{P_n}$
$0 + 0.08 + 0.08 + \sum_{n=0}^{\infty}{P_n}$
$0.16 + \sum_{n=0}^{\infty}{P_n}$

So equivalently we need $\sum_{n=0}^{\infty}{P_n} = 1 - 0.16 = 0.84$

But $P_n$ is a geometric series, so $\sum_{n=0}^{\infty}{P_n} = \frac{a}{1 - r} = \frac{0.08}{1 - r} = 0.84$
$\frac{0.08}{1 - r} = 0.84$
$0.08 = (0.84)(1 - r)$
$0.08 = 0.84 - (0.84)r$
$-0.76 = -(0.84)r$
$r = \frac{-0.76}{-0.84} = \frac{19}{21} \approx 0.9048$

So we calculated $P_n = (0.08)\left(\frac{19}{21}\right)^n$

Finally $P(x) = \begin{array}{ll} 0 & ;\; x = 1 \\ 0.08 & ;\; x =2,3 \\ (0.08)\left( \frac{19}{21} \right)^{x - 4} & ;\; x \geq 4 \end{array}$
• Apr 9th 2013, 11:49 AM
Shakarri
Re: Extrapolation
One inexact way to do it would be to find an equation that describes the probability for 1,2,3 and 4 using the upper bound for your measurements. When you have this approximate equation find the probabilities at 5 to 10, these would be the upper limit of what the probabilities are. Do the same for the lower bound of your measurements. Finding an equation that fits the data should be simple enough because you already have an idea that it is exponential.
This method will tend to give exaggerated upper and lower limits of the probabilities.

Quote:

If the probability is decreasing exponentially, why is it constant at $x = 2,3,4$ ?

I'm pretty sure the $+-8%$ is that he knows the value of the probability lies in the range $p\pm8%$ and there is something special about 0 that means the probability has to be zero and so the uncertainty range is zero.
• Apr 9th 2013, 01:17 PM
StevenDR
Re: Extrapolation
Yeah I meant that at x = 1 I know the probabillity exactly. It is 100% at x = 1
For x = 2, x = 3 and x = 4 I would know the probabillity within a range of +- 8%

for example:
x = 1, 100%
x = 2, 75% +- 8%
x = 3, 50% +- 8%
x = 4, 20% +- 8%

These are just numbers I made up, the reason i'm asking this because I want to know if its worth to measure them since it will take me alot of time to do so. And the higher the x gets the longer it will take. The guess at exponential decrease is just based on the few times I did it without actually measuring anything and on what others tell me.
• Apr 9th 2013, 08:17 PM
Well you can use the same ideas that I used (when I misunderstood your question.) Look at different functions assuming it's a geometric series. BTW, how can $P(x = 1) = 100\%$; then $P(x \neq 0) = 0\%$...?
Do you mean, $P(x >= 1) = 100\%,\; P(x >= 2) = p_2 \pm 8\%,\; \ldots$
In this case: $P(x = 1) = 100\% - (p_2 \pm 8\%)$
I would think about it this way, as computing an infinite sum (of the geometric series) only helps if you assume: $\sum_{x=1}^{\infty}{P(x)} = 100\%$