The mode of a discrete random variable X with p.m.f. p(x) is that value x* for w

Question

The mode of a discrete random variable X with p.m.f. p(x) is that
value x* for which p(x) is largest (the most probable x value).

Consider X distributed poisson(lambda) ... and k positive integer>=1

a) Show that P(X = k-1) < P(X = k) if and only if k < lambda.

{Note:
{Be sure to prove both directions of the "if and only if" in the above!

Thus, the poisson distribution is unimodal, ...
... with the mode occuring at floor(lambda).

b) Show that if lambda is a positive integer then ...
... P(X = lambda) = P(X = lambda - 1).

Thus, if lambda is an integer then both lambda and lambda-1 are modes.

Explanation / Answer

The pmf for the poisson RV is given by P(X = k) = ?^k e^(-?)/k! for k = 0, 1, 2, ... .
---------------
a) P(X = k-1) < P(X = k)
<==> ?^(k-1) e^(-?)/(k-1)! < ?^k e^(-?)/k!
<==> 1 < ?/k
<==> k < ?.
---------------
b) Suppose that ? is an integer.
Then, P(X = ?) = ?^? e^(-?)/?!
........................= ?^(?-1) e^(-?)/(?-1)!, since ?! = ?(? - 1)!
........................= P(X = ?-1).
---------------
c) In this case, P(X = k) = (1-p)^(k-1) * p for k = 1, 2, 3, ..., where p is the probability of a success.

Let f(p) = p(1 - p)^(k-1).
Then, f '(p) = (1 - p)^(k-1) + p * (k-1)(1 - p)^(k-2) * -1
...................= (1 - p)^(k-2) [(1 - p)^1 - p(k - 1)]
...................= (1 - p)^(k-2) (1 - pk).

Assuming that 0 < p < 1, then f '(p) = 0 when k = 1/p.
So, the mode occurs at the value of k (as an integer) closest to 1/p

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