Question

Suppose the random variable X has Poisson distribution with rate parameter l. Let g be the function defined by g(u) = 1/(u+1) , u >0. Show E(g(X)) >g(E(X)).

Answer 1

Answer:

To prove E(g(X)) > g(E(X))

Here it is given that the random variable X has Poisson distribution with rate parameter l

Let us consider LHS

E(g(X))

= E(1 / (1 + x))

= (1 / x+1) e^-I * I^x / x!

= e^-I * I^x / (x+1)!

= e^-I*I^x / (x+1)!

= 1/I e^-I * I^(x+1) / (x+1)!

= 1/I e^-I*I^u / u!

= 1/I [e^-I*I^u / u! - e^-I]

= 1/I (1 - e^-I)

Consider,

E(X) = I

Now g(E(X)) = 1 / (I + 1)

E(g(x)) - g(E(X)) = 1/I - e^-I/I - 1/(1+I)

= 1/(I(I+1) - e^-I/I

= 1/I [ 1/(I+1) - 1/e^I)

[e^I = 1 + I + I^2/2! + ........ >= 1 + I ]

So

1 / (1 + I) > 1/e^I

(1/(1+I) - 1/e^I) > 0

E*g(X)) >= g(E(X)

Here

Therefore , LHS = RHS

Hence it is proved.