Moment-Generating Function

title: Moment-Generating Function (Definition)
The moment generating function (mgf) of a [[Random Variable|r.v.]] $X$ is:  
$$M_X(t) = E(e^{tx})$$
 
- $E(e^{tx})$ is a function of $t$ ($t$ is a dummy variable)
- $E(e^{tx})$ needs to be finite on an open interval containing $0$, otherwise we say the mgf doesn't exist

Ex: Find the mgf for Bernoulli Distribution: $M_X(t)=E(e^{tx})={\sum }_{x\in X}{e}^{tx}(f(x))=e^0(1-p)+e^{t}p$

Ex: Find the mgf of a Uniform Distribution: $M_U(t)=E(e^{tu})={\int }_a^b{e}^{tu}\cdot \frac{1}{b-a}du=\frac{e^{tb}-e^{ta}}{t(b-a)}$

• 1st moment: $E(x)=\mu$
• 2nd moment: $E(x^2)$ → ${\sigma }^2=Var(x)$, Variance
• 3rd moment: $E(x^3)$ → Skewness
• 4th moment: $E(x^4)$ → Kurtosis (how “fat the tail is”)

Why is the mgf important? Because we have the following results.

title: Result 1  
We can calculate the $n$-th moment of a [[Random Variable|r.v.]] $X$ by evaluating the $n$-th derivative of the [[Moment-Generating Function|mgf]] at $t = 0$

MGFs allow us to replace “messy” integration with “cleaner” derivatives.

title: Result 2: The mgf determines the distribution
If two RVs have the same mgf, then they have the same distribution.  

title:Result 3: The mgf of indenpendent distributions
If [[Random Variable|r.v.]]s $X$ and $Y$ are independent, then $M_{X +Y}(t) = M_X(t) \cdot M_Y(t)$
 
This is useful because we are often interested in sums and averages in [[Model|Modelling]].

Example: Find the mgf of a Binomial Distribution, i.e. find the mgf of $X$, where $X\sim Bin(n,p)$. $M_X(t)=E[e^{tx}]=e^{t\cdot 0}\left(\genfrac{}{}{0ex}{}{n}{0}\right)p^0(1-p{)}^n+e^{t-1}(\genfrac{}{}{0ex}{}{n}{1})p^1(1-p{)}^{n-1}+\cdots$ We know that $X=X_1+\cdots +X_n$ where $X_i\sim Ber(p)$

Using Result 3, we get $M_X(t)=M_{X_1}(t)\cdot M_{X_2}(t)\dots M_{X_n}(t)$ $⟹M_X^{\prime }(0)=np\leftarrow E(X)$

TODO: COMPLETE THE PROOF HRE M_1_X WHAT IS happening here????