(3.8) This question introduces a rather efficient method for calculating the mean and variance of probability distributions. We define the moment generating function M(t) for a random variable x by M(t) = (etx). Show that this definition implies that (x) = M(n) (0), (3.51) (3.52) where M(n) (t) mean (x) = d" M/dt" and further that the M (¹) (0) and the variance σ = = M(2)(0) [M(¹) (0)] 2. Hence show that: - (a) for a single Bernoulli trial, = M(t) pe 1-p; (3.53) (b) for the binomial distribution, M(t) = (pe +1 - p)"; (3.54) (c) for the Poisson distribution, M(t) = em(et-1); (3.55) (d) for the exponential distribution, λ M(t) (3.56) Hence derive the mean and variance in each case and show that they agree with the results derived earlier.
(3.8) This question introduces a rather efficient method for calculating the mean and variance of probability distributions. We define the moment generating function M(t) for a random variable x by M(t) = (etx). Show that this definition implies that (x) = M(n) (0), (3.51) (3.52) where M(n) (t) mean (x) = d" M/dt" and further that the M (¹) (0) and the variance σ = = M(2)(0) [M(¹) (0)] 2. Hence show that: - (a) for a single Bernoulli trial, = M(t) pe 1-p; (3.53) (b) for the binomial distribution, M(t) = (pe +1 - p)"; (3.54) (c) for the Poisson distribution, M(t) = em(et-1); (3.55) (d) for the exponential distribution, λ M(t) (3.56) Hence derive the mean and variance in each case and show that they agree with the results derived earlier.
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Chapter1: Units, Trigonometry. And Vectors
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![(3.8) This question introduces a rather efficient method
for calculating the mean and variance of probability
distributions. We define the moment generating
function M(t) for a random variable x by
M(t) = (etx).
Show that this definition implies that
(x) = M(n) (0),
(3.51)
(3.52)
where M(n) (t)
mean (x)
= d" M/dt" and further that the
M (¹) (0) and the variance σ
=
=
M(2)(0) [M(¹) (0)] 2. Hence show that:
-
(a) for a single Bernoulli trial,
=
M(t) pe 1-p;
(3.53)
(b) for the binomial distribution,
M(t) = (pe +1 - p)";
(3.54)
(c) for the Poisson distribution,
M(t) = em(et-1);
(3.55)
(d) for the exponential distribution,
λ
M(t)
(3.56)
Hence derive the mean and variance in each case
and show that they agree with the results derived
earlier.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b3e03bd-95ca-4689-ae67-efa4ab2cc0ad%2Fe46990f3-9360-4497-8a21-6ea867774a56%2Fqwrd3g8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(3.8) This question introduces a rather efficient method
for calculating the mean and variance of probability
distributions. We define the moment generating
function M(t) for a random variable x by
M(t) = (etx).
Show that this definition implies that
(x) = M(n) (0),
(3.51)
(3.52)
where M(n) (t)
mean (x)
= d" M/dt" and further that the
M (¹) (0) and the variance σ
=
=
M(2)(0) [M(¹) (0)] 2. Hence show that:
-
(a) for a single Bernoulli trial,
=
M(t) pe 1-p;
(3.53)
(b) for the binomial distribution,
M(t) = (pe +1 - p)";
(3.54)
(c) for the Poisson distribution,
M(t) = em(et-1);
(3.55)
(d) for the exponential distribution,
λ
M(t)
(3.56)
Hence derive the mean and variance in each case
and show that they agree with the results derived
earlier.
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