Question

Derive the mean and variance of: binomial, exponential,poisson, normal and gammq distribution

Answer 1

Derivation of the Mean and Variance of Binomial distribution:

The mean of the binomial distribution

The mean of the binomial distribution is np.

Hence, mean of the BD is np and the Variance is npq.

Derivation of the Mean and Variance of Exponential distribution:

The mean of the exponential distribution is calculated using the integration by parts.

Hence, the mean of the exponential distribution is 1/λ.

To find the variance of the exponential distribution, we need to find the second moment of the exponential distribution, and it is given by

\(E[X^2] = \int\limits_0^\infty x^2 \lambda e^{-\lambda x} = \frac 2 {\lambda^2}\)

Hence, the variance of the continuous random variable, X is calculated as:

Var(X) = E(X²)- E(X)²

Now, substituting the value of mean and the second moment of the exponential distribution, we get,

Var(X) = \(\frac 2{\lambda^2} - \frac 1 {\lambda^2} = \frac 1{\lambda^2}\)

Thus, the variance of the exponential distribution is 1/λ².

Derivation of the Mean and Variance of Poisson distribution:

Variance (X) = E(X²) - E(X)²

= λ² + λ - (λ)²

= λ

Thus, the variance of the poisson distribution is λ.

Derivation of the Mean and Variance of Gamma distribution:

The mean of the gamma distribution

The mean of the binomial distribution is p.

So, Variance = E[x²] – [E(x²)], where p = (E(x))

= p(p + 1) – p²

= p

Thus, the variance of the gamma distribution is p.