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Determine the distribution that has all moments equal to $\mu$ where $ 0 < \mu < 1$

Having difficulty getting started on this, any help appreciated.

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Note that, in order for the above to be true, we require that, for the distribution's moment generating function $$ \partial_t^nf(t)\bigg|_{t=0} = \mu $$ a simple guess for what this function should be yields $$ f(t) = \mu e^t + C $$ for some arbitrary constant $C$ (this is, in fact, the only analytic function which satisfies the above equality, which can be seen from assuming that $f$ is written as a power series).

Note that this is only true (in the distribution case) whenever we have a probability 'density' (I use this loosely) of the form $$ p(x) = \mu\, \delta(x-1) + (1-\mu)\delta(x-0) $$ which is exactly the Bernoulli distribution.

Guillermo Angeris
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    One may also note the reason that $\mu < 1$ is required. Indeed, as long as $\mu <1$ the unknown distribution must have an exponential moment, and thus its MGF is is defined and analytic on a nonempty interval of the origin. – nullUser Aug 24 '17 at 02:46
  • So, how to rigorously prove "Note that this is only true (in the distribution case) whenever we have a probability 'density' (I use this loosely) of the form"? – Ma Joad Dec 05 '19 at 03:19
  • @Jethro: For $f(t)$ to be a moment generating function, you need $f(0)=1$ and so $C=1-\mu$, i.e. $f(t)=\mu e^t +1-\mu$, which is indeed the mgf of a Bernoulli random variable with parameter (probability or mean) of $\mu$. For the uniqueness issue, see https://math.stackexchange.com/questions/458680/how-to-prove-moment-generating-function-uniqueness-theorem – Henry Dec 05 '19 at 03:31