Question

The first three moments of a distribution about the value 2 of the variables are 1,16, and -40. Show that the mean=3, the variance=15 and ߤଷ = −86.

Answer 1

Step-by-step explanation:

Let rth moment of a variable x about 5 is μ′r=E(xi−5)r and let rth moment of x about its mean be μr=E(xi−x¯)r.

So, μ′1=2⇒E(xi)−5=2⇒E(xi)=x¯=2+5=7

So, first moment about mean = μ1=E(xi−x¯)=E(xi)−x¯=7−7=0...(1)

2nd moment about mean =μ2=E(xi−x¯)2=E(xi−7)2=E[(xi−5)+(5−7)]2=E[(xi−5)−2]2=E(xi−5)2−4E(xi−5)+4=μ′2−4μ′1+4=20−4∗2+4=20−8+4=16....(2)

3rd moment about mean =μ3=E(xi−7)3=E[(xi−5)+(5−7)]3=E[(xi−5)−2]3=E[(xi−5)3−3(x−5)2∗2+3(xi−5)∗22−23]=E(xi−5)3−6E(xi−5)2+12E(xi−5)−8=μ′3−6μ′22+12μ′1−8=40−6∗20+12∗2−8=−64 . .....(3)

4th moment about mean =μ4=E(xi−7)4=E[(xi−5)−(7−5)]4=E(xi−5)4−4C1E(xi−5)3∗2+4C2E(xi−5)2∗(2)2−4C2E(xi−5)∗(2)3+4C4(2)4=μ′4−4∗2μ′3+6∗4μ′2−4∗8μ′1+1∗16=50−8∗40+24∗20−32∗2+1∗16=50−320+480−64+16=546−384=162...(4)

Skewness= √β1=√(μ23μ32)=√(64∗6416∗16∗16)=−1 (negative sign because sign of μ3 is negative)

Kurtosis= β2=μ4μ22=16216∗16∼0.63

Since √β1<0, the distribution is negatively skewed and since β2<3 the distribution is platykurtic

Answer 2

Answer:

Equating the moments we get mean = 3, the variance = 15 and ߤଷ  =  −86.

Step-by-step explanation:

Given :

The first three moments of a distribution about the value 2 of the variables are 1,16, and -40

To find :

mean = 3, the variance = 15 and ߤଷ  =  −86.

Step 1

Let us denote the n-moment about the value 2 by $M_{n, 2}=E\left[(X-2)^{n}\right]$

$M_{1}=E[(X-2)]=1 \ldots(1)$

$M_{2}=E\left[(X-2)^{2}\right]=16 \ldots(2)$

$M_{3}=E\left[(X-2)^{3}\right]=-40 \ldots(3)$

Step 2

From (1), we get

$E[X]-E[2]=1$

$\Rightarrow E[X]-2=1$

$\Rightarrow Mean =\mu_{1}=E[X]=3$

Step 3

From (2), we get

$E\left[(X-2)^{2}\right]=E\left[(X-3+1)^{2}\right]$

$=E\left[(X-3)^{2}+2(X-3)+1\right]$

$=E\left[(X-3)^{2}\right]+2 E(X-3)+E(1)=16$

Step 4

Variance, $\mu_{2}=E\left[(X-3)^{2}\right]$

$=16-2 E[X-3]-E[1]$

$=16-2(E[X]-3)-1$

$=16-2(3-3)-1=\mathbf{1 5}$

Step 5

From (3), we get

$E\left[(X-2)^{3}\right]=E\left[((X-3)+1)^{3}\right]$$=E\left[(X-3)^{3}\right]+3 E\left[(X-3)^{2}\right]+3 E[X-3]+E[1]$$=E\left[(X-3)^{3}\right]+3(15)+3(0)+1$

$=-40$

Step 6

Skewness, $\mu_{3}=E\left[(X-2)^{3}\right]$

$=-40-3(15)-3(0)-1$

$=-86$

Therefore, mean=3, the variance=15 and ߤଷ = −86.

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