I have a function $f(\vec{q})$, which I need to minimize, where $\vec{q} =(q_1,q_2)$. Note that $q_1$ and $q_2$ are probability values. $n_1, n_2, k_1,k_2$ are positive natural numbers. Also, $n_1$ & $n_2$ values are much greater than $k_1$ & $k_2$ values.
\begin{aligned} f(\vec{q}) & \approx\left(n_{1}\right)\left(1-T q_{1}\left(1-q_{1}\right)^{k_{1}}\left(1-q_{2}\right)^{k_{2}}\right) \\ &+\left(n_{2}\right)\left(1-T q_{2}\left(1-q_{1}\right)^{k_{1}}\left(1-q_{2}\right)^{k_{2}}\right) \end{aligned}
I took the 2 partial derivatives and solved them simultaneously to obtain the following: \begin{aligned} q_{1} &=\frac{n_{1}-n_{2} k_{1}+n_{1} k_{2}}{n_{1}\left(k_{1}+k_{2}+1\right)} \\ q_{2} &=\frac{n_{2}-n_{1} k_{2}+n_{2} k_{1}}{n_{2}\left(k_{1}+k_{2}+1\right)} \end{aligned}
Here is the problem I am facing:
From the solutions I got, It seems that $q_1$ or $q_2$ can be negative for some choices of $n_1, n_2, k_1,k_2$, which is not acceptable as $q_1$ or $q_2$ are probability values.
In such cases, how do I find the minimum within the $[0,1] \times [0,1]$ domain. Shall I just check the values of the function at the boundary points. ? I tried that , but all boundary points lead to the same solution $n_1 +n_2$.
Can someone explain how I should go about this?
You can use Lagrange multipliers technique:
$$\left\{\begin{array}{rcl} f(q_{1}, q_{2}) \to \min\limits_{q_{1}, q_{2}} \\ 0 \le q_{1} \le 1 \\ 0 \le q_{2} \le 1 \end{array}\right.$$
$L = f(q_{1}, q_{2}) - \lambda_{1} \cdot q_{1} + \lambda_{2} \cdot (q_{1} - 1) - \lambda_3 \cdot q_{2} + \lambda_4 \cdot (q_{2} - 1)$
And use Karush-Kuhn-Tucker conditions to minimize it:
K.K.T:
$$\left\{\begin{array}{rcl} L^{\prime}_{q_{1}} = 0 \\ L^{\prime}_{q_{2}} = 0 \\ \lambda_{1, 2, 3, 4} \ge 0 \\ \lambda_{1} \cdot q_{1} = 0 \\ \lambda_{2} \cdot (q_{1} - 1) = 0 \\ \lambda_{3} \cdot q_{2} = 0 \\ \lambda_{4} \cdot (q_{2} - 1) = 0 \\ \end{array}\right.$$