Let F(x₁, X₂) = -ln(x₁) - Ln(1 - X₁ - X₂) - Ln(1 - X₂). Is The Function Convex? Find Its Global Minimum Within The Domain, Where (x₁⁰, X₂⁰) = (1/3, 1/3).

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Introduction

In this article, we delve into the analysis of the convexity of the function f(x₁, x₂) = -ln(x₁) - ln(1 - x₁ - x₂) - ln(1 - x₂), and we aim to pinpoint its global minimum within a specified domain. The function presented is a crucial example in optimization, frequently encountered in fields such as economics, engineering, and machine learning. Understanding its properties, particularly convexity, is vital for efficient optimization. Convex functions possess the characteristic that any local minimum is also a global minimum, simplifying the optimization process significantly. Our exploration will involve determining the domain of the function, investigating its convexity by examining its Hessian matrix, and subsequently finding its global minimum. We will also discuss the relevance of the starting point (x₁⁰, x₂⁰) = (1/3, 1/3) in the context of iterative optimization algorithms.

Domain of the Function

Before we begin our investigation into convexity and the global minimum, it is essential to establish the domain of the function. The function f(x₁, x₂) = -ln(x₁) - ln(1 - x₁ - x₂) - ln(1 - x₂) involves natural logarithms. Logarithmic functions are defined only for positive arguments. Therefore, we must ensure that the arguments of all logarithms are strictly greater than zero. This leads to the following conditions:

  1. x₁ > 0
  2. 1 - x₁ - x₂ > 0
  3. 1 - x₂ > 0

These inequalities define the domain of the function. Let's analyze them in detail. The first condition, x₁ > 0, indicates that x₁ must be a positive number. The second condition, 1 - x₁ - x₂ > 0, can be rearranged to x₁ + x₂ < 1. This implies that the sum of x₁ and x₂ must be less than 1. The third condition, 1 - x₂ > 0, rearranges to x₂ < 1, meaning x₂ must be less than 1. Combining these conditions, the domain D of the function f(x₁, x₂) is the set of all points (x₁, x₂) in the two-dimensional space that satisfy all three inequalities simultaneously. Geometrically, this domain is a triangle bounded by the lines x₁ = 0, x₂ = 0, and x₁ + x₂ = 1. This triangular region is an open set, which is important for the subsequent analysis of convexity and the existence of a minimum.

Convexity Analysis

To ascertain whether the function f(x₁, x₂) is convex within its domain, we need to compute the Hessian matrix and analyze its properties. The Hessian matrix is a square matrix of second-order partial derivatives of a function. For a two-variable function like ours, the Hessian matrix H is given by:

H =
\begin{bmatrix}
  \frac{\partial^2 f}{\partial x_1^2} & \frac{\partial^2 f}{\partial x_1 \partial x_2} \\
  \frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2^2}
\end{bmatrix}

A function is convex if its Hessian matrix is positive semi-definite everywhere in its domain. This means that all the eigenvalues of the Hessian matrix are non-negative, or equivalently, that all the leading principal minors of the Hessian matrix are non-negative.

First, we compute the first-order partial derivatives:

\frac{\partial f}{\partial x_1} = -\frac{1}{x_1} + \frac{1}{1 - x_1 - x_2}
\frac{\partial f}{\partial x_2} = \frac{1}{1 - x_1 - x_2} + \frac{1}{1 - x_2}

Next, we compute the second-order partial derivatives:

\frac{\partial^2 f}{\partial x_1^2} = \frac{1}{x_1^2} + \frac{1}{(1 - x_1 - x_2)^2}
\frac{\partial^2 f}{\partial x_2^2} = \frac{1}{(1 - x_1 - x_2)^2} + \frac{1}{(1 - x_2)^2}
\frac{\partial^2 f}{\partial x_1 \partial x_2} = \frac{\partial^2 f}{\partial x_2 \partial x_1} = \frac{1}{(1 - x_1 - x_2)^2}

Now, we can form the Hessian matrix:

H =
\begin{bmatrix}
  \frac{1}{x_1^2} + \frac{1}{(1 - x_1 - x_2)^2} & \frac{1}{(1 - x_1 - x_2)^2} \\
  \frac{1}{(1 - x_1 - x_2)^2} & \frac{1}{(1 - x_1 - x_2)^2} + \frac{1}{(1 - x_2)^2}
\end{bmatrix}

To determine if the Hessian is positive semi-definite, we need to check the leading principal minors. The first leading principal minor is the top-left element:

D_1 = \frac{1}{x_1^2} + \frac{1}{(1 - x_1 - x_2)^2}

Since x₁ > 0 and 1 - x₁ - x₂ > 0 in the domain, D₁ is always positive. The second leading principal minor is the determinant of the entire Hessian matrix:

det(H) = \left(\frac{1}{x_1^2} + \frac{1}{(1 - x_1 - x_2)^2}\right) \left(\frac{1}{(1 - x_1 - x_2)^2} + \frac{1}{(1 - x_2)^2}\right) - \left(\frac{1}{(1 - x_1 - x_2)^2}\right)^2

Simplifying the determinant:

det(H) = \frac{1}{x_1^2(1 - x_1 - x_2)^2} + \frac{1}{x_1^2(1 - x_2)^2} + \frac{1}{(1 - x_1 - x_2)^2(1 - x_2)^2}

Since x₁, 1 - x₁ - x₂, and 1 - x₂ are all positive in the domain, det(H) is also always positive. Since both leading principal minors are positive, the Hessian matrix is positive definite, and therefore, the function f(x₁, x₂) is strictly convex in its domain.

Finding the Global Minimum

Since the function f(x₁, x₂) is strictly convex within its domain, any local minimum is also the global minimum. To find the global minimum, we need to find the critical points by setting the first-order partial derivatives equal to zero:

\frac{\partial f}{\partial x_1} = -\frac{1}{x_1} + \frac{1}{1 - x_1 - x_2} = 0
\frac{\partial f}{\partial x_2} = \frac{1}{1 - x_1 - x_2} - \frac{1}{1 - x_2} = 0

From the first equation, we have:

\frac{1}{x_1} = \frac{1}{1 - x_1 - x_2}

which implies:

x_1 = 1 - x_1 - x_2

2x_1 + x_2 = 1

From the second equation, we have:

\frac{1}{1 - x_1 - x_2} = \frac{1}{1 - x_2}

which implies:

1 - x_1 - x_2 = 1 - x_2

x_1 = 0

However, x₁ = 0 contradicts the domain condition x₁ > 0. Let's correct the second equation. The original second equation is:

\frac{\partial f}{\partial x_2} = -\frac{1}{1 - x_1 - x_2} + \frac{1}{1 - x_2} = 0

Thus,

\frac{1}{1 - x_1 - x_2} = \frac{1}{1 - x_2}

This implies:

1 - x_2 = 1 - x_1 - x_2

Which simplifies to:

x_1=0

Which is not possible.

Let's re-examine the equations:

-\frac{1}{x_1} + \frac{1}{1 - x_1 - x_2} = 0 	 (1)
-\frac{1}{1 - x_1 - x_2} + \frac{1}{1 - x_2} = 0 	 (2)

From equation (1):

\frac{1}{x_1} = \frac{1}{1 - x_1 - x_2}

x_1 = 1 - x_1 - x_2 	 (3)

From equation (2):

\frac{1}{1 - x_1 - x_2} = \frac{1}{1 - x_2}

1 - x_1 - x_2 = 1 - x_2

x_1 = 0

Again we got x₁=0, which is not feasible. Let's resolve this again. The second equation should actually be:

\frac{\partial f}{\partial x_2} = -\frac{1}{1 - x_1 - x_2} + \frac{1}{1 - x_2} = 0

This gives us:

\frac{1}{1 - x_1 - x_2} = \frac{1}{1 - x_2}

Which leads to:

1 - x_2 = 1 - x_1 - x_2

Simplifying, we get x₁ = 0, but this contradicts x₁ > 0. Thus, there must be an error in our earlier equations. Adding equations (1) and (2), we get:

-\frac{1}{x_1} + \frac{1}{1 - x_2} = 0

So,

\frac{1}{x_1} = \frac{1}{1 - x_2}

Which gives:

x_1 = 1 - x_2 	 (4)

Substituting (4) into (3):

1 - x_2 = 1 - (1 - x_2) - x_2

1 - x_2 = x_2

2x_2 = 1

x_2 = \frac{1}{2}

Then,

x_1 = 1 - x_2 = 1 - \frac{1}{2} = \frac{1}{2}

Thus, the critical point is (1/2, 1/2). Since 1 - x₁ - x₂ = 1 - 1/2 - 1/2 = 0, this point is not in the domain. This indicates there's no global minimum within the domain. The function will approach negative infinity as (x₁, x₂) approaches the boundary.

Role of the Starting Point (x₁⁰, x₂⁰) = (1/3, 1/3)

The starting point (x₁⁰, x₂⁰) = (1/3, 1/3) is within the domain D, as 1/3 > 0, 1 - 1/3 > 0, and 1 - 1/3 - 1/3 = 1/3 > 0. This point is relevant when using iterative optimization algorithms such as gradient descent or Newton's method to find the minimum. Since the function is strictly convex, these algorithms, if implemented correctly, should converge to the global minimum if it existed within the domain. However, as we've determined that the global minimum occurs on the boundary (where the function tends to negative infinity), these algorithms will move towards the boundary without converging to a specific point within the domain. The starting point influences the path taken by the algorithm but will not change the fundamental behavior of moving towards the boundary.

Conclusion

In conclusion, the function f(x₁, x₂) = -ln(x₁) - ln(1 - x₁ - x₂) - ln(1 - x₂) is strictly convex within its defined domain. While a critical point was calculated to be (1/2, 1/2), it falls outside the domain where 1 - x₁ - x₂ > 0. Thus, there is no global minimum within the domain; the function decreases without bound as it approaches the boundaries of the domain. The starting point (x₁⁰, x₂⁰) = (1/3, 1/3) is a feasible point within the domain and would serve as a valid starting point for iterative optimization methods, although these methods will ultimately steer towards the boundary rather than converging to a minimum inside the domain.