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Python Program to calculate function from indicator random variable from given condition
Suppose we have two values k and n. Consider a random permutation say p1, p2, ..., pn of first n natural numbers numbers 1, 2, ..., n and calculate the value F, such that F = (X2+...+Xn-1)k, where Xi is an indicator random variable, which is 1 when one of the following two conditions holds: pi-1 < pi > pi+1 or pi-1 > pi < pi+1 and Xi is 0 otherwise. We have to find the expected value of F.
So, if the input is like k = 1 n = 1000, then the output will be 1996/3
To solve this, we will follow these steps −
- Define a function exp_factor() . This will take n,k
- if k is same as 1, then
- return(2*(n-2) , 3)
- otherwise when k is same as 2, then
- return (40*n^2 -144*n + 131, 90)
- otherwise when k is same as 3, then
- return (280*n^3 - 1344*n^2 +2063*n -1038,945)
- otherwise when k is same as 4, then
- return (2800*n^4 - 15680*n^3 + 28844*n^2 - 19288*n + 4263, 14175)
- otherwise when k is same as 5, then
- return (12320*n^5 - 73920*n^4 + 130328*n^3 - 29568*n^2 - 64150*n -5124, 93555)
- return 1.0
- From the main method, do the following −
- M := n-2
- p := 2.0/3
- q := 1 - p
- (num, den) := exp_factor(n, k)
- g := gcd(num, den)
- return fraction (num/g) / (den/g)
Example
Let us see the following implementation to get better understanding −
from math import gcd def exp_factor(n,k): if k == 1: return (2*(n-2),3) elif k == 2: return (40*n**2 -144*n + 131,90) elif k == 3: return (280*n**3 - 1344*n**2 +2063*n -1038,945) elif k == 4: return (2800*n**4 - 15680*n**3 + 28844*n**2 - 19288*n + 4263, 14175) elif k == 5: return (12320*n**5 - 73920*n**4 + 130328*n**3 - 29568*n**2 - 64150*n -5124, 93555) return 1.0 def solve(k, n): M = n-2 p = 2.0/3 q = 1 - p num, den = exp_factor(n,k) g = gcd(num, den) return str(int(num/g))+'/'+str(int(den/g)) k = 1 n = 1000 print(solve(k, n))
Input
1, 1000
Output
1996/3
Updated on: 11-Oct-2021
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