Evaluation of Risk in Investments

Given two investment options A and B, we have tho find the less risky investment of the two. The two investments A and B are each represented by an array. Each element in the array is a probable investment outcome. Thus each element in the array is a pair of two values. The first value is the amount of money received and the second value is the probability that this money can be received. For instance if A = [ (100,0.1), (200,0.2) (300,0.7) ], it means that there is 10 % probability to earn Rs 100, 20% probability to earn Rs 200 and 70% chance to earn Rs 300 from investment A.

We have to use a statistical approach to solve the problem. For each investment, we first calculate an average amount of money that can be earned from it. Secondly, we also calculate the standard deviation in the money earned. Then we need to normalize this standard deviation by dividing it by the mean.

Each probable outcome is an observation. The probability for each amount of money is its frequency. Since the observations are given with frequencies we need to apply the following formulas to calculate the mean and standard deviation



If X denotes the set of observations (x_i,f_i).
Mean = \bar X = \sum{(x_i * f_i)} / \sum{f_i}
Standard deviation  = \sigma = \sqrt{\sigma^2} = \sum{((x_i - \bar X)^2*f_i}) / \sum{f_i}

Let us take an example to demonstrate how to apply this method.
Example

Input:  A = [(0,0.1), (100,0.1), (200,0.2), (333,0.3), (400,0.3) ]
        B = [ (100,0.1), (200,0.5), (700,0.4) ]

Explanation:
Mean Investment of A
Index | Outcome | Probability | Probability*Outcome
(i)       (xi)        (fi)        xi*fi
----------------------------------------------------------
1          0          0.1            0
2        100          0.1           10
3        200          0.2           40
4        333          0.3         99.9
5        400          0.3          120
----------------------------------------------------------
Total:                1.0        269.1
Mean = 269.1/1 = 269.1

Mean Investment of B:
Index | Outcome | Probability | Probability*Outcome
(i)       (xi)        (fi)        xi*fi
----------------------------------------------------------
1        100          0.1           10
2        200          0.5          100
3        700          0.4          280
----------------------------------------------------------
Total:                1.0          390
Mean = 390/1 = 390.1

Standard Deviation of A
Mean = 269.1
Index | Outcome | Probability | (xi-Mean)^2 | A*fi
(i)       (xi)        (fi)        (A)
----------------------------------------------------------
1          0          0.1         72414.81  7241.481 
2        100          0.1         28594.81  2859.481
3        200          0.2          4774.81   954.962
4        333          0.3          4083.21  1224.963
5        400          0.3         17134.81  5140.443
----------------------------------------------------------
Total:                1.0                   17421.33
Standard Deviation  = sqrt(17421.33/1) = 131.989
Normalized Standard Deviation = 131.989/269.1 = 0.49

Standard Deviation of B
Mean = 390.1
Index | Outcome | Probability | (xi-Mean)^2 | A*fi
(i)       (xi)        (fi)        (A)
----------------------------------------------------------
1        100          0.1         84158.01   8415.801
2        200          0.5         36138.01  18069.005
3        700          0.4         96100.00  38440.000
----------------------------------------------------------
Total:                1.0                   64924.801
Standard Deviation  = sqrt(64924.801/1) = 254.803
Normalized Standard Deviation: 254.803 / 390.1 = 0.65

Since Investment A has lesser normalized standard deviation,
it is less risky.


Input: A = [(0,0.1), (100,0.1), (200,0.2), (333,0.3), (400,0.3) ]
       B = [ (100,0.1), (200,0.5), (700,0.4) ]

Explanation: 
For Investment A
Average: 269.9
Standard Deviation: 131.987
Normalised Std: 0.489024
For Investment B
Average: 258.333
Standard Deviation: 44.8764
Normalised Std: 0.173715
Investment B is less risky

The CPP implementation of the problem is given below

C++

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// C++ code for above approach
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
  
// First Item in the pair is the
// value of observation (xi).
// Second Item in the pair is 
// the frequency of xi (fi)
typedef pair<float,float> Data;
  
// Vector stores the observation 
// in pairs of format (xi, fi), 
// where xi = value of observation
typedef vector<Data> Vector;
  
// Function to calculate the 
// summation of fi*xi
float sigma_fx(const Vector & v)
{
    float sum = 0;
    for ( auto i : v) {
        sum += i.first * i.second; 
    }
    return sum;
}
  
// Function to calculate summation fi
float sigma_f(const Vector & v)
{
    float sum = 0.0;
    for ( auto i : v) {
        sum += i.second;
    }
    return sum;
}
  
// Function to calculate the mean
// of the set of observations v
float calculate_mean(const Vector & v)
{
    return sigma_fx(v) / sigma_f(v);
}
  
// Function to calculate the std
// deviation of set of observations v
float calculate_std(const Vector & v) 
{
    // Get sum of frequencies
    float f = sigma_f(v);
      
    // Get the mean of the set 
    // of observations
    float mean = sigma_fx(v) / f;
      
    float sum = 0;
      
    for (auto i: v) {
        sum += (i.first-mean)*
               (i.first-mean)*i.second;
    }
      
    return sqrt(sum/f);
}
  
// Driver Code
int main() 
{
      
    Vector A = { {0,0.1}, {100,0.1}, 
               {200,0.2}, {333,0.3}, {400,0.3}};
    Vector B = { {100,0.1}, {200,0.5}, {700,0.4}};
  
    float avg_A = calculate_mean(A);
    float avg_B = calculate_mean(B);
    float std_A = calculate_std(A);
    float std_B = calculate_std(B);
      
      
    cout << "For Investment A" << endl;
    cout << "Average: " << avg_A << endl;
    cout << "Standard Deviation: " << 
                           std_A << endl;
    cout << "Normalised Std: " << 
                    std_A / avg_A << endl;
    cout << "For Investment B" << endl;
    cout << "Average: " << avg_B << endl;
    cout << "Standard Deviation: " << 
                            std_B << endl;
    cout << "Normalised Std: " << std_B / 
                            avg_B << endl;
      
    (std_B/avg_B) < (std_A/avg_A) ? cout << 
            "Investment B is less risky\n":
            cout << "Investment A is less risky\n";
      
    return 0;
}

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Output:

For Investment A
Average: 269.9
Standard Deviation: 131.987
Normalised Std:  0.489024
For Investment B
Average: 390
Standard Deviation: 254.755
Normalised Std:  0.653217
Investment A is less risky

References
https://www.statcan.gc.ca/edu/power-pouvoir/ch12/5214891-eng.htm
std::accumulate cppreference.com



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