I have the following constrained optimization problem corresponding to the maximum likelihood density estimation: \begin{equation} \begin{aligned} &\text{maximize} && L(f) \\ &\text{subject to} && f \in H \\ &&& \int_a^b f(x) \mathop{}\!\mathrm{d} x = 1 \\ &&& f(x) \geq 0 \text{ for all } x \in [a,b]. \end{aligned} \end{equation} where $x$ is a random variable with probability density function (PDF) $f$ on an interval $[a,b] \subset \textrm{IR}$, and $H$ is a subspace of $L^1 [a,b]$ (i.e., Lebesgue integrable on $[a,b]$).
I need to construct some PDFs $f_n$ to prove the existence of a solution to the above optimization problem, which have the following properties:
- Continuous and positive on the interval $(-1,1)$,
- Integrates to one on the interval $[-1,1]$,
- Vanishes at $(-1)$ and $1$,
- Equal to $n$ at $x=0$ (e.g., $f_2=2$ at $x=0$).
These functions $f_n$ are graphically represented in the figure below. My question is how to mathematically represent the functions $f_n$.
Thanks.
Warning: This solution may be a bit overkill. Also, the pdfs below do not really "look like" the pdfs you search for, e.g. they are for example not differentiable. But maybe it still helps to have a concrete family that satisfies the properties you described.
If you just need $f_n$ to be continuous, you could use the family of pdfs $h_r \ (r \in [1, \infty))$ where
$$h_r \colon [-1,1] \to [0,\infty), x \mapsto \begin{cases} r\cdot (1 -r\cdot |x|), & |x| \leq \frac{1}{r} \\ 0 , & \text{otherwise}. \end{cases}$$
For every $r \geq 1$, we have that $h_r$ is continuous, even piecewise differentiable. Furthermore, it's non-negative, positive on $(-1/r, 1/r)$ and as its area is made out of two triangles of height $r$ and width $1/r \leq 1$, the integral is one. It's also clear that $h_r(0) = r$.
Now, note that the linear combination of two pdfs is again a pdf. So, let's just do the following:
For $n = 1$, just take $f_1= h_1$.
For $n \geq 2$, as $f_n$ should be positive on $(-1, 1)$, let one of the pdfs involved be $h_1$ and its weight be $1/n$.
Now, it suffices to find $r \geq 1$ such that for $$f_n(x) = \frac{1}{n} h_1(x) + \frac{n-1}{n} h_r(x)$$ we have $f_n(0) = n$. Plugging in gives
$$n \overset{!}{=} \frac{1}{n} + \frac{n-1}{n} \cdot r \implies r = \left(n - \frac{1}{n}\right)\cdot \frac{n}{n-1} = n+1.$$
So
$$f_n(x) = \frac{1}{n} h_1(x) + \frac{n-1}{n} h_{n+1}(x)$$
does the trick.