Re: P=e^-λ((λ^2)/2+λ+1)
- From: "David Park" <djmpark@xxxxxxxxxxx>
- Date: Mon, 7 Jul 2008 11:27:06 -0400
I hope I have your precedence correct. With Mathematica
Define the lambda function so we can look at it:
f[\[Lambda]_] := Exp[-\[Lambda] (\[Lambda]^2/2 + \[Lambda] + 1)]
Plot it:
Plot[f[\[Lambda]], {\[Lambda], 0, 5}, PlotRange -> {0, 1}]
We might try numerical root finding, but t Mathematica will find the three roots with Solve and we can extract the real one.
sol = Part[Solve[f[\[Lambda]] == p, \[Lambda]], 1, 1] // Quiet
\[Lambda] ->
1/3 (-2 -
2/(10 - 27 Log[p] + 3 Sqrt[3] Sqrt[4 - 20 Log[p] + 27 Log[p]^2])^(
1/3) + (10 - 27 Log[p] +
3 Sqrt[3] Sqrt[4 - 20 Log[p] + 27 Log[p]^2])^(1/3))
Checking visually for a value of p:
sol /. p -> .5
\[Lambda] -> 0.447766
which looks about right on the plot.
--
David Park
djmpark@xxxxxxxxxxx
http://home.comcast.net/~djmpark/
<davidhaslam100@xxxxxxxxxxxxxx> wrote in message news:74532254-ad81-4fb7-b1d0-c65c959291e7@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Hi,
Can anyone explain how I might solve the above equation for λ? This
relates to the probability of a soccer match ending under 2.5 goals
(given an average expected amount of goals (λ). I'm stuck.
Thanks, David
.
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