Transform methods  (Page 3/5)

Some absolutely continuous distributions

1. Uniform on $(a,b)$ ${\displaystyle f_X\left(t\right)=\frac{1}{b-a}a<t<b}$
   $$M_X\left(s\right)=\int e^{st}{f}_X\left(t\right)dt=\frac{1}{b-a}{\int }_a^b{e}^{st}dt=\frac{e^{sb}-e^{sa}}{s(b-a)}$$
2. Symmetric triangular $(-c,,c)$
   $$f_X\left(t\right)=I_{[-c,0)}\left(t\right)\frac{c+t}{c^2}+I_{[0,c]}\left(t\right)\frac{c-t}{c^2}$$
   $$M_X\left(s\right)=\frac{1}{c^2}{\int }_{-c}^0(c+t)e^{st}dt+\frac{1}{c^2}{\int }_0^c(c-t)e^{st}dt=\frac{e^{cs}+e^{-cs}-2}{c^2{s}^2}$$
   $$=\frac{e^{cs}-1}{cs}·\frac{1-e^{-cs}}{cs}=M_Y\left(s\right)M_Z(-s)=M_Y\left(s\right)M_{-Z}\left(s\right)$$
   where M _Y is the moment generating function for $Y\sim$ uniform $(0,c)$ and similarly for M _Z . Thus, X has the same distribution as the difference of two independent random variables, each uniform on $(0,c)$ .
3. Exponential $\left(\lambda \right)$ $f_X\left(t\right)=\lambda e^{-\lambda t},t\ge 0$
   In example 1, above, we show that ${\displaystyle M_X\left(s\right)=\frac{\lambda }{\lambda -s}}$ .
4. Gamma $(\alpha ,\lambda )$ ${\displaystyle f_X\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\lambda }^{\alpha }{t}^{\alpha -1}{e}^{-\lambda t}t\ge 0}$
   $$M_X\left(s\right)=\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{t}^{\alpha -1}{e}^{-(\lambda -s)t}dt={\left[\frac{\lambda }{\lambda -s}\right]}^{\alpha }$$
   For $\alpha =n$ , a positive integer,
   $$M_X\left(s\right)={\left[\frac{\lambda }{\lambda -s}\right]}^n$$
   which shows that in this case X has the distribution of the sum of n independent random variables each exponential $\left(\lambda \right)$ .
5. Normal $(\mu ,{\sigma }^2)$ .
   • The standardized normal, $Z\sim N(0,1)$
     $$M_Z\left(s\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{st}{e}^{-t^2/2}dt$$
     Now ${\displaystyle st-\frac{t^2}{2}=\frac{s^2}{2}-\frac{1}{2}{(t-s)}^2}$ so that
     $$M_Z\left(s\right)=e^{s^2/2}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-{(t-s)}^2/2}dt=e^{s^2/2}$$
     since the integrand (including the constant $1/\sqrt{2\pi }$ ) is the density for $N(s,1)$ .
   • $X=\sigma Z+\mu$ implies by property (T1)
     $$M_X\left(s\right)=e^{s\mu }{e}^{{\sigma }^2{s}^2/2}=exp\left(\frac{{\sigma }^2{s}^2}{2},+,s,\mu \right)$$

Moment generating function and simple random variables

Suppose ${\displaystyle X=\sum _{i=1}^n{t}_i{I}_{A_i}}$ in canonical form. That is, A _i is the event $\{X=t_i\}$ for each of the distinct values in the range of X , with $p_i=P\left(A_i\right)=P(X=t_i)$ . Then the moment generating function for X is

The moment generating function M _X is thus related directly and simply to the distribution for random variable X .

Consider the problem of determining the sum of an independent pair $\{X,Y\}$ of simple random variables. The moment generating function for the sum is the product ofthe moment generating functions. Now if ${\displaystyle Y=\sum _{j=1}^m{u}_j{I}_{B_j}}$ , with $P(Y=u_j)={\pi }_j$ , we have

The various values are sums $t_i+u_j$ of pairs $(t_i,u_j)$ of values. Each of these sums has probability $p_i{\pi }_j$ for the values corresponding to $t_i,u_j$ . Since more than one pair sum may have the same value, we need to sort the values,consolidate like values and add the probabilties for like values to achieve the distribution for the sum. We have an m-function mgsum for achieving this directly. It produces the pair-products for the probabilities and the pair-sums forthe values, then performs a csort operation. Although not directly dependent upon the moment generating function analysis, it produces the same result as that produced by multiplyingmoment generating functions.