Suppose that X is a poisson random variable with parameter l

Suppose that X is a poisson random variable with parameter lambda=0.4. That is, for an integer k, P(X = k) = e^0.4(0.4)^k/k! . Use Markov\'s inequality to find an upper bound for P(X greterthan or equal to 1). Then calculate exactly P(X greterthan or equal to 1). How does this compare to the upper bound in part a)

Solution

a)

Note that

P(x>=a) <= E(x) / a

Thus, as E(x) = 0.4, and a = 1,

P(x>=1) <= 0.4/1

P(x>=1) <= 0.4 [ANSWER]

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b)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    0.4      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.670320046

As

P(x>=1) = 1 - P(0),

P(x>=1) = 1 - 0.670320046 = 0.329679954 [ANSWER]

Indeed, this is less than the upper bound in a). [ANSWER]