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Theoretical Distribution

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Question : 114 of 118

Marks: +1, -0

If

x

is a Poisson variate such that

P(x=2)=

9 P(x=4)+90 P(x=6),

find mean of x

$m=2$
$m=1$
$m=\pm 1$
$m=-4$

Solution:

Concept:

Use Poisson probability formula

P(x=k)=\frac{e^{-m}m^{k}}{k!}

and equate given condition to solve for mean

m

.

Explanation:

Given:

P(x=2)=9P(x=4)+90P(x=6)

.

Substitute Poisson formula:

\frac{e^{-m}m^{2}}{2!}=9\cdot\frac{e^{-m}m^{4}}{4!}+90\cdot\frac{e^{-m}m^{6}}{6!}

.

Cancel

e^{-m}\neq0

.

\frac{m^{2}}{2}=9\cdot\frac{m^{4}}{24}+90\cdot\frac{m^{6}}{720}

.

Simplify fractions:

9/24=3/8

,

90/720=1/8

.

So

\frac{m^{2}}{2}=\frac{3}{8}m^{4}+\frac{1}{8}m^{6}

.

Multiply both sides by 8:

4m^{2}=3m^{4}+m^{6}

.

Bring all terms:

m^{6}+3m^{4}-4m^{2}=0

.

Factor:

m^{2}(m^{4}+3m^{2}-4)=0

.

Let

y=m^{2}

:

y^{2}+3y-4=0 \Rightarrow (y+4)(y-1)=0 \Rightarrow y=1

or

y=-4

.

Since

y=m^{2}\ge0

,

m^{2}=1 \Rightarrow m=\pm1

.

For Poisson distribution, mean

m>0

, so

m=1

.

Answer:

m=1

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