Respuesta :
Answer:
a) Mean, E(X) = - 0.5
Variance = = 9.25
b) [tex]M_{X}(t)=E(e^{tX})[/tex]
or
⇒ [tex]=\sum e^{tx}p(x)[/tex]
⇒ [tex]=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}[/tex]
Step-by-step explanation:
Given:
moment generating function of X as:
MX(t) = [tex]\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}[/tex]
a) Now
Mean, E(X) = [tex]M_{X}'(t=0)[/tex]
Thus,
[tex]M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}[/tex]
or
[tex]M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}[/tex]
also,
[tex]E(X^{2})=M_{X}''(t=0)[/tex]
Thus,
[tex]M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}[/tex]
or
[tex]M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}[/tex]
Therefore,
Mean, E(X) = [tex]M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}[/tex]
or
Mean, E(X) = - 0.5
and
[tex]E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}[/tex]
or
[tex]E(X^{2})[/tex] = 9.5
also,
Variance(X) = E(X²) - E(X)²
⇒ 9.5 - (-0.5)²
= 9.25
b) Now,
Let f(x) be the PMF of X
Thus,
[tex]M_{X}(t)=E(e^{tX})[/tex]
or
⇒ [tex]=\sum e^{tx}p(x)[/tex]
⇒ [tex]=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}[/tex]
Therefore,
at x = 0, P(x) = [tex]\frac{1}{2}[/tex]
at x= - 4 ,P(x) = [tex]\frac{1}{3}[/tex]
at x = 5, P(x) = [tex]\frac{1}{6}[/tex]
Thus,
E(X) =[tex]\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})[/tex]
or
E(X) = - 0.5
also,[tex]E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})[/tex]
[tex]E(X^{2})[/tex] = 9.5
Hence,
Var(X) = E(X²) - E(X)²
⇒ 9.5 - (-0.5)²
= 9.25
