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I came across a question which asked to obtain the probability function of $X$ (a discrete random variable) with its characteristic function given as follows: $${\phi _X}(t) = {e^{\lambda ({e^{it}} - 1)}}$$

I know that this is the characteristic function of a Poisson Distribution. So, $$X \sim Poi(\lambda )$$

However, I was unable to show this mathematically. I started to answer this question as follows: $$P(X = x) = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {{e^{ - itx}}{\phi _X}(t)dt} $$ $$ = \frac{1}{{2\pi }}{e^{ - \lambda }}\int\limits_{ - \pi }^\pi {{e^{ - itx}}{e^{\lambda {e^{it}}}}dt} $$

But after this step I am unable to figure how will be evaluate the integral.

Can someone please suggest how should go ahead after this step?

Thanks in advance!

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The formula I used (see exercise $26.12$, Probability and Measure by Patrick Billingsley), similar to the celebrated inversion formula, is (the formula that you gave can be derived from it): $$P[X = a] = \lim_{T \to \infty} \frac{1}{2T}\int_{-T}^T e^{-ita}\phi_X(t) dt.$$ Notice that $\phi_X(t) = e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k e^{itk}}{k!}$. Evaluate the integral by Fubini's theorem \begin{align} & \int_{-T}^T e^{-ita}\phi_X(t) dt \\ = & \int_{-T}^T e^{-ita}e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k e^{itk}}{k!} dt \\ = & e^{-\lambda}\sum_{k = 0}^\infty\frac{\lambda^k}{k!}\int_{-T}^Te^{it(k - a)}dt \\ = & 2Te^{-\lambda}\frac{\lambda^a}{a!} + 2e^{-\lambda}\sum_{k \neq a}\frac{\lambda^k\sin[(k - a)T]}{k!(k - a)} \end{align} where we used that if $k = a$, then $\int_{-T}^T e^{it(k - a)} dt = 2T$, and if $k \neq a$, $$\int_{-T}^T e^{it(k - a)} dt = 2\int_0^T \cos[(k - a)t] dt = \frac{2}{k - a}\sin[(k - a)T].$$ Notice by the dominated convergence theorem, $$\lim_{T \to \infty}\frac{1}{2T}\sum_{k \neq a}\frac{\lambda^k\sin[(k - a)T]}{k!(k - a)} = \sum_{k \neq a}\lim_{T \to \infty}\frac{\lambda^k\sin[(k - a)T]}{2k!(k - a)T} = 0.$$ Therefore, $P[X = a] = e^{-\lambda}\frac{\lambda^a}{a!}$, for $a = 0, 1, 2, \ldots$, the proof is complete.

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    $\begingroup$ Thank you for the solution. The formula for the Inversion Theorem that I used was one the two formulas that our probability teacher had told us we could use in case of Discrete Random Variables. The other one was the one that you have used in the solution. However, using the former formula, I was able to derive the PMF of the Binomial Distribution from its Characteristic Function. $\endgroup$ – Kaustav Sen Nov 7 '15 at 8:46
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    $\begingroup$ I see, during my derivation, I found your formula should be OK too. And you can reach the answer similarly as in my answer. Just expand $e^{\lambda e^{it}}$ to power series and integrate term by term. In your formula the integration when $k \neq x$ will be exactly $0$ due to periodicity of $\sin$ function. $\endgroup$ – Zhanxiong Nov 7 '15 at 14:03
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Given cf ${\phi _X}(t) = {e^{\lambda ({e^{it}} - 1)}}$, derive the pmf $P(X=x)$ ...

There is a simpler way. For a discrete random variable $X$ taking non-negative integer values, the cf $E[e^{i t X}]$ is related to the probability generating function pgf $E[s^{X}]$ via $s = e^{i t}$, so the pgf has form:

$$\Pi(s) = E\left[s^X\right] = e^{\lambda (s-1)}$$

The pgf generates probabilities via the relation:

$$P(X=x) \quad = \quad \frac{1}{x!}\frac{d^x\Pi(s)}{ds^x}|_{s=0} \quad \text{for} \quad x\in \{0,1,2,\ldots \}$$

The $n^{th}$ derivative of $e^{\lambda (s-1)}$ wrt $s$ has form $\lambda ^n e^{\lambda (s-1)}$, and when $s = 0$, the latter has form: $e^{-\lambda } \lambda ^n$. Replacing $n$ with $x$, the pmf is thus:

$$P(X=x) = \frac{e^{-\lambda } \lambda ^x}{x!} \quad \text{for} \quad x\in \{0,1,2,\ldots \}$$

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