Respuesta :
Answer:
a.) X  ≈ N (0, 12)
b.) Y ≈  exp(2)
c.) The given distribution is not defined.
d.) Not a standard form of any distribution .
Step-by-step explanation:
Given - In parts (a)−(d) below, either use the information given to determine
      the distribution of the random variable, or show that the information
      given is not sufficient by describing at least two different random
      variables that satisfy the given condition.
To find - (a) X is a random variable such that Mâ‚“(t) = [tex]e^{6t^{2} }[/tex] when |t| < 2.
       (b) Y is a random variable such that  [tex]M_{Y}[/tex](t) = [tex]\frac{2}{2 - t}[/tex] for t < 0.5.
       (c) Z is a random variable such that [tex]M_{Z}[/tex](t) = ∞ for t ≥ 5.
       (d) W is a random variable such that [tex]M_{W}[/tex](2) = 2
Proof -
a.)
As we know that the moment generating function of normal distribution is given by
Mâ‚“(t) = [tex]e^{ut + \frac{1}{2}t^{2} \sigma^{2}}[/tex]
Now,
Mâ‚“(t) = [tex]e^{6t^{2} }[/tex]
    = [tex]e^{0(t) + \frac{12}{2} t^{2} }[/tex]
⇒μ = 0, σ = 12
Nd it is represented by
X  ≈ N (0, 12)
b.)
Moment generating function of exponential function is given by
[tex]M_{Y}[/tex](t) = [tex]( 1 - \frac{t}{\theta})^{-1}[/tex]
Now,
 [tex]M_{Y}[/tex](t) = [tex]\frac{2}{2 - t}[/tex] for t < 0.5.
     = [tex]\frac{1}{\frac{2 - t}{2} }[/tex]
     = [tex]\frac{1}{1 - \frac{t}{2} }[/tex]
     = [tex]( 1 - \frac{t}{2} )^{-1}[/tex]
And it is represented by
Y ≈  exp(2)
c.)
Z is a random variable such that [tex]M_{Z}[/tex](t) = ∞ for t ≥ 5.
The given distribution is undefined.
d.)
[tex]M_{W}[/tex](2) = E([tex]e^{2w}[/tex] )
      = [tex]\int\limits^{-\infty}_{\infty} {e^{2w} f(w) } \, dw[/tex]
     = 2
It is not the standard form of any distribution.
