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Mathematics

1 answer:

Serjik [45]2 years ago

6 0

Answer:

Step-by-step explanation:

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Define a new random variable by y = 2px. show that, as p l 0, the mgf of y converges to that of a chi squared random variable wi

Mandarinka [93]

The moment generating function (mgf) of y is given by

$M_y(t)=E(e^{ty})=E(e^{2pxt})=M_x(2pt)=p(1-e^{2pt}(1-p))^{-1}$

Since the momoent generating function of x is given by $M_x(t)=p(1-e^{t}(1-p))^{-1}$

When p tends to 0, we have

$\lim_{p \to 0} M_y(t)= \lim_{p \to 0} \frac{p}{1-e^{2pt}(1-p)} = \frac{0}{0}$

Applying L'Hopital's rule we have:

$\lim_{p \to 0} M_y(t)=\lim_{p \to 0} \frac{1}{e^{2pt}+2te^{2pt}+2pte^{2pt}} = \frac{1}{1+2t} , \ \ \ t\ \textless \ \frac{1}{2}$

This shows that y converges to <span>a chi squared random variable with 2r degrees of freedom</span>.

5 0

2 years ago

Graphing Exponential Function in Exercise ,sketch the graph of the function.See example 3 and 4.

boyakko [2]

Answer:

graphic attached

Step-by-step explanation:

when t = 0 then s (t) = 1/4

when t tends to infinity then s (t) tends to 0

when t tends to infinity then s (t) tends to infinity

the function has a horizontal asymptote at y = 0

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