please tell me a way to solve this problem and tell me the following answer which is my own is correct
The Mode Of Binomial Random Variable
A binomial random variable denotes the number of observations of a sequence of independent bernoulli
trials, when each trial is mutually exclusive and has the same probability of attaining the success. The
formula for the Binomial probability density function is,
P(X=x) = ncxpxqn-x ; x=0,1,2,3,............,n
where n is a positive integer & 0
Here n is the number of trials concerned and p is the probability of a trial attaining the success.
The mode of an X random value is the x value which maximises the value of its probability mass
function.
In order to find the mode we need to consider the ratio of successive probabilities as follows,
P(X=x) = ncxpxqn-x ; x=0,1,2,3,.........,n
and
P(X=x-1) = ncx-1px-1qn-x+1 ; x=1,2,3,.........,n
Thus by considering the ratio of the two expressions we get,
P(X=x)/P(X=x-1) = ncxpxqn-x/ ncx-1px-1qn-x+1 = {(n-x+1)/x}(p/q) = {(n+1)/x-1}(p/q)
Then the largest x integer which satisfies the following condition can be taken as the mode of the
Binomial random variable.
P(X=x)/P(X=x-1) > 1 or P(X=x) > P(X=x-1)
{(n+1)/x-1}(p/q) > 1
By simplifying we get
x < (n+1)p
Thus let's consider the largest integer which satisfies this condition.
Case-1
When (n+1)p is an integer x = (n+1)p satisfies the condition.
So P[X=(n+1)p] = P[X=(n+1)p-1]
\ x = (n+1)p & x = (n+1)p-1 are both modes.
Case-2
Then (n+1)p is not an integer the mode is the largest integer which is less than (n+1)p.
Note:-
Please share your ideas and comments about this article with me by sending me an email to
bhindunil.sl@gmail.com
The Mode Of Binomial Random Variable
A binomial random variable denotes the number of observations of a sequence of independent bernoulli
trials, when each trial is mutually exclusive and has the same probability of attaining the success. The
formula for the Binomial probability density function is,
P(X=x) = ncxpxqn-x ; x=0,1,2,3,............,n
where n is a positive integer & 0
The mode of an X random value is the x value which maximises the value of its probability mass
function.
In order to find the mode we need to consider the ratio of successive probabilities as follows,
P(X=x) = ncxpxqn-x ; x=0,1,2,3,.........,n
and
P(X=x-1) = ncx-1px-1qn-x+1 ; x=1,2,3,.........,n
Thus by considering the ratio of the two expressions we get,
P(X=x)/P(X=x-1) = ncxpxqn-x/ ncx-1px-1qn-x+1 = {(n-x+1)/x}(p/q) = {(n+1)/x-1}(p/q)
Then the largest x integer which satisfies the following condition can be taken as the mode of the
Binomial random variable.
P(X=x)/P(X=x-1) > 1 or P(X=x) > P(X=x-1)
{(n+1)/x-1}(p/q) > 1
By simplifying we get
x < (n+1)p
Thus let's consider the largest integer which satisfies this condition.
Case-1
When (n+1)p is an integer x = (n+1)p satisfies the condition.
So P[X=(n+1)p] = P[X=(n+1)p-1]
\ x = (n+1)p & x = (n+1)p-1 are both modes.
Case-2
Then (n+1)p is not an integer the mode is the largest integer which is less than (n+1)p.
Note:-
Please share your ideas and comments about this article with me by sending me an email to
bhindunil.sl@gmail.com
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What you've done is quite right.
Just do the same for Poisson distribution: P(x) = k^x * e^(-k) / x!
In fact, it is much easier.
To find the value of X which maximizes P(X), P(x=X)/P(x=X-1) > 1 ..... or P(x=X) > P(x=X-1)
=> k^x * e^(-k) / x! > k^(x-1) * e^(-k) / (x-1)!
=> k>x OR x
Thus the mode for Poisson variable is k if k is an integer, and largest integer less than k, if k is not an integer
Just do the same for Poisson distribution: P(x) = k^x * e^(-k) / x!
In fact, it is much easier.
To find the value of X which maximizes P(X), P(x=X)/P(x=X-1) > 1 ..... or P(x=X) > P(x=X-1)
=> k^x * e^(-k) / x! > k^(x-1) * e^(-k) / (x-1)!
=> k>x OR x
Thus the mode for Poisson variable is k if k is an integer, and largest integer less than k, if k is not an integer