Question

(a) For a distribution, the mean is 10, variance is 16, the skewness sk₄ is +1 and kurtosis b₂ is 4. Obtain the first four moments about the origin i.e. zero. Comment upon the nature of the distribution. 

 (b) Find the mean and variance of binomial distribution.

Answer 1

$Question \space \\ (A) for \space a \space distribution \space , the \space mean \space is \space 10, variance \space is \space 16,coefficient \space of \space skewness \space is \space +1 \space and \space coefficient \space of \space kurtosis \space is \space 4. Obtain \space the \space first \space tour \space moment \space about \space the \space origine \space i.e.,zero. \\ Comment \space upon \space the \space nature \space of \space distribution \space .\\ Solution \\ Given \\ Mean=m'_1=10, variance=m'_2=16, coefficient \space of \space skewness \space =\gamma_1=1, coefficient \space of \space kurtosis \space is \space \beta_2=4. \\ 1st \space moment \space about \space the \space origine \space is \space \\ Mean=m'_1=10\\ Second \space moment \space about \space the \space origine \space is \space \\ m'_2=variance+(\frac{\sum x }{n })^{2 } =16+10 ^{2 }=116 \\ Coefficient \space of \space skewness \space is \space +1 \\ =\gamma _1 =+1 = \frac{m _3 }{m ^{\frac{3 }{2 } } _2 }\\ m ^{ \frac{3 }{2 }} _2 = 16 ^{ \frac{3 }{ 2 } } = 64 \\ Coefficient \space of \space kurtosis \space is \space 4. \\ \beta = 4 \\ \frac{m _4 }{ m ^{2 }_2 } = 4 \\ m _ 4 = 4(16)^2 = 1024 \\ Third \space moment \space about \space the \space origine \space is \space \\ m'_3 =m _ 3 + 3m _2 m' _1 +m _1 ^{3 } \\ 64+3(16)(10)+10^3 = 1544\\ Fourth \space moment \space about \space the \space origin \space is \\ m'_4 =m_4 + 4 m_3 m'_1 + 6 m_2 m'_2 + (m'_1) ^4= 1024+4(64)(10)+6(16)10^2+10^4=23184\\ It \space is \space a \space positive \space skew \space distribution \\ . \\ . \\$

$(B)Let\space X\space be\space a\space random\space variable\space \\with\space the\space Binomial\space distribution.\space \space The\space probability\space function\space of\space X\space is\\ p(x)=\dbinom{n}{x}p^x q^{n-x}\\ then\\ E(X)=\displaystyle\sum_{i=0}^n x.\dbinom{n}{x}p^x q^{n-x}\\ =\displaystyle\sum_{i=0}^n x.\frac{n!}{(n-x)!x!}p^x q^{n-x}\\ =\displaystyle\sum_{i=1}^n \frac{n!}{(n-x)!(x-1)!}p^x q^{n-x}\\ (since\space the \space x=0 \space term \space vanishes)\\ E(X)=\displaystyle\sum_{i=1}^n \frac{n(n-1)!}{(n-x)!(x-1)!}(p)p^{x-1} q^{n-x}\\ =np\displaystyle\sum_{i=1}^n \frac{(n-1)!}{(n-x)!(x-1)!}p^{x-1} q^{n-x}\\ =np\displaystyle\sum_{i=1}^n \frac{(n-1)!}{[(n-1)-(x-1)]!(x-1)!}p^{x-1} q^{n-x}\\ =np\displaystyle\sum_{i=1}^n \dbinom{n-1}{x-1} p^{x-1} q^{n-x}\\ =np[^{n-1}C_0 q^{n-1}+^{n-1}C_1 pq^{n-2}+^{n-1}C_2 p^2q^{n-3}+...+^{n-1}C_{n-1} p^{n-1}]\\ =np[this \space is \space binomial\space expansion\space of\space (p+q)^{n-1}]\\ =np[(p+q)^{n-1}]\\ but\space we\space know\space that \space (p+q)=1\\ so E(X)=np(1)^{n-1}=np\\ thus \space the \space mean \space of\space binomail\space distribution\space is\space np.\\ now \\ Var (X)=E(X^2)-[E(X)]^2\\ E(X^2)=\displaystyle\sum_{i=0}^nx^2. \dbinom{n}{x}p^x q^{n-x}\\ E(X^2)=\displaystyle\sum_{i=0}^n[x(x-1)+x]. \dbinom{n}{x}p^x q^{n-x}\\ E(X^2)=\displaystyle\sum_{i=0}^n[x(x-1)]. \dbinom{n}{x}p^x q^{n-x}+\sum_{i=0}^nx. \dbinom{n}{x}p^x q^{n-x}\\ E(X^2)=\displaystyle\sum_{i=2}^n \frac{x(x-1)n!}{(n-x)!x(x-1)(x-2)!}p^x q^{n-x}+np\\ E(X^2)=n(n-1)p^2\\ \displaystyle\sum_{i=2}^n \frac{(n-2)!}{(n-x)!(x-2)!}p^{x-2} q^{n-x}+np\\ E(X^2)=n(n-1)p^2\displaystyle\sum_{i=2}^n \frac{(n-2)!}{[(n-2)-(x-2)]!(x-2)!}p^{x-2} q^{n-x}+np\\ E(X^2)=n(n-1)p^2\displaystyle\sum_{i=2}^n \dbinom{n-2}{x-2} p^{x-2} q^{n-x}+np\\ =n(n-1)p^2[^{n-2}C_0 q^{n-2}+^{n-2}C_1 pq^{n-3}+...+^{n-2}C_{n-2} p^{n-2}]+np\\ =n(n-1)p^2[(p+q)^{n-2}]+np\\ since \\ p+q=1, we \space have \\ E(X^2)=n(n-1)p^2+np\\ using \space this,\\ Var(X)=n(n-1)p^2+np-(np)^2\\ =n^2p^2-np^2+np-n^2p^2\\ =np(1-p)=npq\\ hence \space the \space variance \space of \space binomial\space distribution \space is \space npq.$