Mathematical Statistics: Very Difficult Problems with Solutions

Solution:
For one observation
[tex]P(X=x)=\binom{n}{x}p^{x}(1-p)^{n-x}[/tex]
Since the observations are independent, the likelihood function is
[tex]L(p)=\prod_{i=1}^{k}\binom{n}{x_i}p^{x_i}(1-p)^{n-x_i}=C\,p^{S}(1-p)^{nk-S}[/tex]
where [tex]S=x_1+x_2+\ldots+x_k[/tex] and [tex]C=\prod_{i=1}^{k}\binom{n}{x_i}[/tex] does not depend on [tex]p[/tex]. Let [tex]0\lt S\lt nk[/tex]. For [tex]0\lt p\lt 1[/tex] we take the logarithm and set its derivative equal to zero:
[tex]\ln L(p)=\ln C+S\ln p+(nk-S)\ln(1-p)[/tex]
[tex]\frac{d}{dp}\ln L(p)=\frac{S}{p}-\frac{nk-S}{1-p}=0\iff S(1-p)=(nk-S)p\iff p=\frac{S}{nk}[/tex]
The second derivative [tex]-\frac{S}{p^2}-\frac{nk-S}{(1-p)^2}[/tex] is negative, so this critical point is the maximum. The same formula holds in the boundary cases: for [tex]S=0[/tex] we have [tex]L(p)=C(1-p)^{nk}[/tex], which is largest at [tex]p=0[/tex], and for [tex]S=nk[/tex] we have [tex]L(p)=Cp^{nk}[/tex], which is largest at [tex]p=1[/tex]. Hence
[tex]\hat{p}=\frac{x_1+x_2+\ldots+x_k}{nk}=\frac{\bar{x}}{n}[/tex]
i.e. the relative frequency of successes in all [tex]nk[/tex] trials.

Solution:
Let [tex]r[/tex] be the correlation coefficient. The regression line of [tex]Y[/tex] on [tex]X[/tex] is [tex]y-\bar{y}=b_{Y|X}(x-\bar{x})[/tex] with slope [tex]b_{Y|X}=r\frac{\sigma_Y}{\sigma_X}[/tex], and the regression line of [tex]X[/tex] on [tex]Y[/tex] is [tex]x-\bar{x}=b_{X|Y}(y-\bar{y})[/tex] with slope [tex]b_{X|Y}=r\frac{\sigma_X}{\sigma_Y}[/tex].
We solve the first given line for [tex]Y[/tex]; the second one is already solved for [tex]X[/tex]:
[tex]Y=1.5+0.1X\Rightarrow b_{Y|X}=0.1, \qquad X=2.5+0.15Y\Rightarrow b_{X|Y}=0.15[/tex]
Check: [tex]r^2=b_{Y|X}\times b_{X|Y}=0.015\le 1[/tex]. If the roles of the two lines were swapped, we would get [tex]r^2=10\times \frac{1}{0.15}\approx 66.7\gt 1[/tex], which is impossible, so the given assignment is the only possible one. Dividing the two slopes (the factor [tex]r\ne 0[/tex] cancels), we get
[tex]\frac{b_{X|Y}}{b_{Y|X}}=\frac{\sigma_X^2}{\sigma_Y^2}=\frac{D(X)}{D(Y)}\Rightarrow \frac{D(X)}{D(Y)}=\frac{0.15}{0.1}=\frac{3}{2}[/tex]

Solution:
We compute the theoretical moments using [tex]\int_{0}^{\infty}x^{s-1}e^{-\lambda x}\,dx=\frac{\Gamma(s)}{\lambda^{s}}[/tex] and [tex]\Gamma(s+1)=s\,\Gamma(s)[/tex]:
[tex]EX=\frac{\lambda^{p}}{\Gamma(p)}\times \frac{\Gamma(p+1)}{\lambda^{p+1}}=\frac{p}{\lambda}, \qquad EX^2=\frac{\lambda^{p}}{\Gamma(p)}\times \frac{\Gamma(p+2)}{\lambda^{p+2}}=\frac{p(p+1)}{\lambda^2}[/tex]
[tex]DX=EX^2-(EX)^2=\frac{p(p+1)}{\lambda^2}-\frac{p^2}{\lambda^2}=\frac{p}{\lambda^2}[/tex]
The sample mean and the second central sample moment are
[tex]\bar{x}=\frac{1+2+6}{3}=3, \qquad m_2=\frac{1^2+2^2+6^2}{3}-3^2=\frac{41}{3}-9=\frac{14}{3}[/tex]
The method of moments equates the theoretical moments with the sample ones:
[tex]\frac{p}{\lambda}=3, \qquad \frac{p}{\lambda^2}=\frac{14}{3}[/tex]
Dividing the first equation by the second, we get
[tex]\lambda=\frac{3}{14/3}=\frac{9}{14}, \qquad p=3\lambda=\frac{27}{14}[/tex]
In general [tex]\hat{\lambda}=\frac{\bar{x}}{m_2}[/tex] and [tex]\hat{p}=\frac{\bar{x}^2}{m_2}[/tex]. By the law of large numbers [tex]\bar{x}[/tex] and [tex]m_2[/tex] converge in probability to [tex]EX[/tex] and [tex]DX\gt 0[/tex], and the estimates are continuous functions of them, so they are consistent. Hence [tex]\hat{\lambda}=\frac{9}{14}[/tex] and [tex]\hat{p}=\frac{27}{14}[/tex].

Solution:
Let [tex]x_i[/tex] be the yields of variety A ([tex]n_1=6[/tex]) and [tex]y_j[/tex] those of variety B ([tex]n_2=5[/tex]). The sample means are
[tex]\bar{x}=\frac{528}{6}=88, \qquad \bar{y}=\frac{510}{5}=102[/tex]
The sums of squared deviations from the means are
[tex]\sum_{i=1}^{6}(x_i-88)^2=22^2+18^2+21^2+7^2+24^2+12^2=2018[/tex]
[tex]\sum_{j=1}^{5}(y_j-102)^2=20^2+14^2+14^2+16^2+4^2=1064[/tex]
Since the variances are assumed equal, we use the pooled estimate of the variance:
[tex]s^2=\frac{2018+1064}{n_1+n_2-2}=\frac{3082}{9}\approx 342.44[/tex]
Under the hypothesis [tex]H_0:\ \mu_1=\mu_2[/tex] the statistic [tex]t=\frac{\bar{x}-\bar{y}}{s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex] has Student's [tex]t[/tex]-distribution with [tex]n_1+n_2-2=9[/tex] degrees of freedom. We compute
[tex]s\sqrt{\frac{1}{6}+\frac{1}{5}}=\sqrt{342.44\times \frac{11}{30}}\approx\sqrt{125.56}\approx 11.21[/tex]
[tex]t=\frac{88-102}{11.21}\approx -1.25\Rightarrow |t|\approx 1.25[/tex]
Since [tex]|t|\approx 1.25\lt 2.262[/tex], the hypothesis [tex]H_0[/tex] is not rejected: at significance level [tex]0.05[/tex] the data do not show a difference in the mean productivity of the two varieties (the difference of [tex]14[/tex] pounds between the sample means can be explained by random variation).

Solution:
First we find the least-squares line. The means are
[tex]\bar{v}=\frac{10+11+12+13+14}{5}=12, \qquad \bar{x}=\frac{5100+4400+4000+3400+3100}{5}=4000[/tex]
The deviations of [tex]v[/tex] from the mean are [tex]-2, \ -1, \ 0, \ 1, \ 2[/tex] and those of [tex]x[/tex] are [tex]1100, \ 400, \ 0, \ -600, \ -900[/tex], hence
[tex]\sum(v_i-\bar{v})(x_i-\bar{x})=-2200-400+0-600-1800=-5000, \qquad \sum(v_i-\bar{v})^2=4+1+0+1+4=10[/tex]
The least-squares coefficients are
[tex]a=\frac{\sum(v_i-\bar{v})(x_i-\bar{x})}{\sum(v_i-\bar{v})^2}=\frac{-5000}{10}=-500, \qquad b=\bar{x}-a\bar{v}=4000+500\times 12=10000[/tex]
so the demand function is [tex]x(v)=10000-500v[/tex]. The revenue is [tex]v\,x(v)[/tex] and the costs are [tex]15000+4x(v)[/tex], so the profit is
[tex]P(v)=(v-4)(10000-500v)-15000=-500v^2+12000v-55000[/tex]
From [tex]P'(v)=-1000v+12000=0[/tex] we get [tex]v=12[/tex], and since [tex]P''(v)=-1000\lt 0[/tex], this is the maximum. At this price the print run is [tex]x(12)=4000[/tex] copies, and the maximal profit (in dollars) is
[tex]P(12)=(12-4)\times 4000-15000=17000[/tex]

Solution:
Let [tex]N=\binom{n}{k}[/tex] and number the samples [tex]j=1,\ 2,\ \ldots,\ N[/tex]. Let [tex]S_j[/tex] be the set of objects in sample [tex]j[/tex] and [tex]\bar{x}_j=\frac{1}{k}\sum_{i\in S_j}x_i[/tex] its mean. The required average is
[tex]G=\frac{1}{N}\sum_{j=1}^{N}\bar{x}_j=\frac{1}{kN}\sum_{j=1}^{N}\sum_{i\in S_j}x_i[/tex]
We change the order of summation by counting, for each object [tex]i[/tex], how many samples contain it. A sample containing object [tex]i[/tex] is determined by choosing its other [tex]k-1[/tex] members from the remaining [tex]n-1[/tex] objects, so object [tex]i[/tex] belongs to exactly [tex]\binom{n-1}{k-1}[/tex] samples. Hence
[tex]\sum_{j=1}^{N}\sum_{i\in S_j}x_i=\binom{n-1}{k-1}\sum_{i=1}^{n}x_i[/tex]
Moreover,
[tex]\binom{n-1}{k-1}=\frac{(n-1)!}{(k-1)!\,(n-k)!}=\frac{k}{n}\times \frac{n!}{k!\,(n-k)!}=\frac{k}{n}N[/tex]
Therefore
[tex]G=\frac{1}{kN}\times \frac{k}{n}N\sum_{i=1}^{n}x_i=\frac{1}{n}\sum_{i=1}^{n}x_i=\bar{x}[/tex]
which is what we had to prove. For example, for daily wages [tex]55, \ 50, \ 52, \ 48, \ 50[/tex] (mean [tex]51[/tex]) the average of the means of the [tex]10[/tex] samples of size [tex]2[/tex] is again [tex]51[/tex]; equal values of different workers count as different objects.