Find the continuous function defined over $[0,1]$ such that $\int_0^1f(x){\rm d}x=1, \int_0^1 xf(x){\rm d}x=a$ and $\int_0^1x^2f(x){\rm d}x=a^2 .$

Question

Find the continuous function defined over $[0,1]$ such that $$\int_0^1 x^kf(x){\rm d}x=a^k,$$ where $k=0,1,2.$

My Thought

Indeed, the conditions imply that $$\int_0^1f(x)=1,~~~\int_0^1 xf(x){\rm d}x=a,~~~\int_0^1 x^2f(x){\rm d}x=a^2.$$ Thus, it's easy to obtain that $$\begin{align*}\int_0^1(1-x)^2f(x){\rm d}x&=\int_0^1(1-2x+x^2)f(x){\rm d}x\\&=\int_0^1f(x){\rm d}x-2\int_0^1xf(x){\rm d}x+\int_0^1 x^2f(x){\rm d}x\\&=1-2a+a^2\\&=(1-a)^2\end{align*}.$$ But how to go on with this? I'm stuck here. Please offer some help. Thanks.

Answer 1

There are infinitely many functions with this property. If you are trying to find one such function take $f(x)$ to be a polynomial of degree 2 and solve for the coefficients from the given conditions.

Answer 2

Since $$ \int_0^1 {f(x)dx} = 1 $$ then f(x) - if assumed to be non-negative - is a probability distribution function that is defined by its moments.
Your problem goes under the Hausdorff moment problem class.

You do not specify what the moments higher than $2$ shall be, so let's examine two different cases.

a) case $k = 0..\infty$

Let's consider first the case in which the all the moments are specified to be equal to $a^k$.

Then given $$ \int_0^1 {x^{\,k} f(x)dx} = a^{\,k} $$ it is also true that $$ {{t^{\,k} } \over {k!}}\int_0^1 {x^{\,k} f(x)dx} = \int_0^1 {{{t^{\,k} x^{\,k} } \over {k!}}f(x)dx} = {{t^{\,k} a^{\,k} } \over {k!}} $$ and that $$ \int_0^1 {\sum\limits_{0\, \le \,k} {{{t^{\,k} x^{\,k} } \over {k!}}} f(x)dx} = \int_0^1 {e^{\,t\,x} f(x)dx} = \sum\limits_{0\, \le \,k} {{{t^{\,k} a^{\,k} } \over {k!}}} = e^{\,a\,t} $$

Since your function is defined in $[0,1]$, then we can write $$ \int_0^1 {e^{\,t\,x} f(x)dx} = \int_{ - \infty }^\infty {e^{\,t\,x} f(x)\left( {H(x) - H(x - 1)} \right)dx} = e^{\,a\,t} $$ where $H(x)$ is the Heaviside function, and the expression is the Moment Generating function or Laplace Transform of $f(x)$ restricted to the $[0,1]$ range and null outside.

Without going on in that approach, if we compute instead the central moments $$ \eqalign{ & \overline x = \int_0^1 {xf(x)dx} = a \cr & \mu _{\,k} = \int_0^1 {\left( {x - \overline x } \right)^{\,k} f(x)dx} = \int_0^1 {\sum\limits_{0\, \le \,j\, \le \,k} {\left( { - 1} \right)^{\,k - j} \left( \matrix{ k \cr j \cr} \right)\overline x ^{\,k - j} x^{\,j} } f(x)dx} = \cr & = \sum\limits_{0\, \le \,j\, \le \,k} {\left( { - 1} \right)^{\,k - j} \left( \matrix{ k \cr j \cr} \right)\overline x ^{\,k - j} \int_0^1 {x^{\,j} f(x)dx} } = \cr & = \sum\limits_{0\, \le \,j\, \le \,k} {\left( { - 1} \right)^{\,k - j} \left( \matrix{ k \cr j \cr} \right)a^{\,k} } = 0^{\,k} a^{\,k} \cr} $$ we discover that they are all null.

That means that, provided $0 \le a \le 1$, the only non-negative $f(x)$ having the required moments is the Dirac delta function $\delta(x-a)$.

b) case higher moments null

Assuming that the moments higher than $2$ are null, we can always proceed as above.

But if you are looking to express $f(x)$ as a polynomial, waiving on the requirement of being non-negative, then we can follow a shorter way.

Putting $$ f(x) = \sum\limits_{0\, \le \,n\, \le \,h} {c_{\,n} x^{\,n} } $$ we get $$ a^{\,k} = \int_0^1 {x^{\,k} f(x)dx} = \int_0^1 {\sum\limits_{0\, \le \,n\, \le \,h} {c_{\,n} x^{\,n + k} } dx} = \sum\limits_{0\, \le \,n\, \le \,h} {{{c_{\,n} } \over {n + k + 1}}} $$ that is $$ \left( {\matrix{ 1 & {1/2} & {1/3} & \cdots \cr {1/2} & {1/3} & {1/4} & \cdots \cr {1/3} & {1/4} & {1/5} & \cdots \cr \vdots & \vdots & \vdots & \ddots \cr } } \right)\left( {\matrix{ {c_{\,0} } \cr {c_{\,1} } \cr {c_{\,2} } \cr \vdots \cr } } \right) = \left( {\matrix{ 1 \cr a \cr {a^{\,2} } \cr \vdots \cr } } \right) $$

The matrix is invertible, although I could not find an easy expression for its inverse.
For a limited number of factors, we can invert it numerically and for $h=2$ we obtain for instance

$f(x)=(180a^2 - 180a + 30)x^2 + (192a - 180a^2 - 36)x + (30a^2 - 36a + 9)$

Answer 3

Let $$ f(x)=b_2x^2+b_1x+b_0. $$ Then $$ \int_0^1f(x)dx=\frac13b_2+\frac12b_1+b_0, \int_0^1xf(x)dx=\frac14b_2+\frac13b_1+\frac12b_0$$ and $$ \int_0^1x^2f(x)dx=\frac15b_2+\frac14b_1+\frac13b_0. $$ Let $$ \frac13b_2+\frac12b_1+b_0=1,\frac14b_2+\frac14b_1+\frac12b_0=a,\frac15b_2+\frac14b_1+\frac13b_0=a^2 $$ and then $$ b_0=3 \left(10 a^2-12 a+3\right),b_1=-12 \left(15 a^2-16 a+3\right),b_2=30 \left(6 a^2-6 a+1\right). $$ So $$ f(x)=30 \left(6 a^2-6 a+1\right)x^2-12 \left(15 a^2-16 a+3\right)x+3 \left(10 a^2-12 a+3\right).$$