Question

5.30. Let ( left{X_{n} ight}_{n=1}^{infty} ) be a sequence of binomial random variables where the probability mass function of ( X_{n}(n=1,2, ldots) ) is given by [ p_{n}(k)=inom{n}{k} p^{k}(1-p)^{n-k}, quad k=0,1,2, ldots, n, ] where ( 0<p<1 ). Further, let the random variable ( Y_{n} ) be defined as [ Y_{n}=frac{X_{n}}{n}-p ] (a) Show that ( Eleft(X_{n} ight)=n p ) and ( operatorname{Var}left(X_{n} ight)=n p(1-p) ). (b) Apply Chebyshev's inequality to show that [ Pleft(left|Y_{n} ight| geq epsilon ight) leq frac{p(1-p)}{n epsilon^{2}}, ] where ( epsilon>0 ). (c) Deduce from (b) that ( Y_{n} ) converges in probability to zero. [Note: This result is known as Bernoulli's law of large numbers.]

          5.30. Let ( left{X_{n}
ight}_{n=1}^{infty} ) be a sequence of binomial random variables where the probability mass function of ( X_{n}(n=1,2, ldots) ) is given by
[
p_{n}(k)=inom{n}{k} p^{k}(1-p)^{n-k}, quad k=0,1,2, ldots, n,
]
where ( 0<p<1 ). Further, let the random variable ( Y_{n} ) be defined as
[
Y_{n}=frac{X_{n}}{n}-p
]
(a) Show that ( Eleft(X_{n}
ight)=n p ) and ( operatorname{Var}left(X_{n}
ight)=n p(1-p) ).
(b) Apply Chebyshev's inequality to show that
[
Pleft(left|Y_{n}
ight| geq epsilon
ight) leq frac{p(1-p)}{n epsilon^{2}},
]
where ( epsilon>0 ).
(c) Deduce from (b) that ( Y_{n} ) converges in probability to zero. [Note: This result is known as Bernoulli's law of large numbers.]

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$5.30. Let ( leftXn ightn=1^infty ) be a sequence of binomial random variables where the probability mass function of ( Xn(n=1,2, ldots) ) is given by [ pn(k)=inomnk p^k(1-p)^n-k, quad k=0,1,2, ldots, n, ] where ( 0<p<1 ). Further, let the random variable ( Yn ) be defined as [ Yn=fracXnn-p ] (a) Show that ( Eleft(Xn ight)=n p ) and ( operatornameVarleft(Xn ight)=n p(1-p) ). (b) Apply Chebyshev's inequality to show that [ Pleft(left|Yn ight| geq epsilon ight) leq fracp(1-p)n epsilon^2, ] where ( epsilon>0 ). (c) Deduce from (b) that ( Yn ) converges in probability to zero. [Note: This result is known as Bernoulli's law of large numbers.]$

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Introduction to Mathematical Statistics

Robert V. Hogg, Joseph W. McKean, Allen… 8th Edition

Chapter 5

Instant Answer

Solved by Expert Kirsty Gledhill

05/30/2024

Step 1

This is a well-known result from the theory of binomial distributions. The variance of a binomial random variable is given by \(\operatorname{Var}(X_n) = np(1-p)\). This is also a well-known result from the theory of binomial distributions. Show more…

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00:01 Okay, so we first want to show that the expectation of xn is equal to np.

00:07 So the expectation of xn is given by the sum of the probability mass function summed over all the values that x can take, which is from k equals 0 to n, times that value.

00:28 So that's where we add on an extra k here.

00:31 Now, that is the same as the sum from k equals 1 to n, because when k equals 0 this is clearly going to make the whole thing vanish.

00:45 And then we can check that this is the same as the sum from k equals 1 to n of n times n minus 1 choose k minus 1 times p to the k times 1 minus p to the n minus k.

01:06 And that is equal to np times the sum from k equals 1 to n of n minus 1 choose k minus 1 times p to the k minus 1 times 1 minus p to the n minus 1 minus k minus 1.

01:31 And that is just exactly the same as np times the sum from j equals 0 to n minus 1 of n minus 1, choose j, p to the j, 1 minus p to the n minus 1 minus j.

01:51 That's just saying j equals k minus 1 and we can see this is the binomial formula for p plus 1 minus p to the power n minus 1 which is just 1 to the n minus 1 which is just 1 and so that gives us np.

02:11 Now i'm not going to show it because it takes an awful long time well i mean it took a long time to show that one it doesn't actually take too long to do but it's just very similar to what we've just done and so the expectation of xn squared you can find is n squared p squared plus np times one minus p, and therefore the variance of xn, which we know is defined as the expectation of xn squared minus the expectation of xn squared, is given by n squared p squared plus np one minus p minus n squared p squared, which is just going to give us np times one minus p.

02:50 Then we can say that chebyshev's inequality tells us that the probability that the difference between xn and its mean being bigger than k standard deviations is less than or equal to 1 over k squared...

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