wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is .


  1. 1

Open in App
Solution

The correct option is A 1
nth moment of a Random variable is given as

E[xn]=xnfx(x)dx

Now, xx¯x; ¯x= mean value of Random Variable x

E[[x¯x]n]=(x¯x)nfx(x)dx

Now, if n=2 [second moment]

E[(x¯x)2]=(x¯x)2fx(x)dx=σ2x

(Variance) [as per definition]

Hence, Variance of a random variable:

σ2x=E[x2]E2[x] ...(i)

Let mean value of poisson-distributed random variable x is λ and we know that Mean = Variance.

In case of Poisson distribution,

So λ=2λ2 (using (i))

λ2+λ2=0

λ=1,2

λ2

λ=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Poisson Distribution
ENGINEERING MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon