Respuesta :
Answer:
(a) moment generating function for X is [tex]\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)[/tex]
(b) [tex]\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}[/tex]
Step-by step explanation:
Given X represents the number on die.
The possible outcomes of X are 1, 2, 3, 4, 5, 6.
For a fair die, [tex]P(X)=\frac{1}{6}[/tex]
(a) Moment generating function can be written as [tex]M_{x}(t)[/tex].
[tex]M_x(t)=\sum_{x=1}^{6} P(X=x)[/tex]
[tex]M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}[/tex]
[tex]M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)[/tex]
(b) Now, find [tex]E(X) \text { and } E\((X^{2})[/tex] using moment generating function
[tex]M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)[/tex]
[tex]M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)[/tex]
[tex]\Rightarrow E(X)=\frac{21}{6}[/tex]
[tex]M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)[/tex]
[tex]M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)[/tex]
[tex]\Rightarrow E\left(X^{2}\right)=\frac{91}{6}[/tex]
Hence, (a) moment generating function for X is [tex]\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)[/tex].
(b) [tex]\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}[/tex]
