Question

If the m.g.f. of a random variable X is\(\psi \left( t \right) = {e^{{t^2}}}\,for - \infty < t < \infty \)What is the distribution of X?

Short answer

Distribution of X is normal distribution with\(\mu = 0\)and\({\sigma ^2} = 2\)

Step-by-step solution

Given information

Mgf of a random variable is\[\psi \left( {\rm{t}} \right) = {{\rm{e}}^{{t^2}}}\,{\rm{for}} - \infty < {\rm{t}} < \infty \]

Calculating distribution of X

Mgf of normal distribution is

\({M_x}\left( t \right) = \exp \left\{ {\mu t + \frac{{{\sigma ^2}{t^2}}}{2}} \right\}\)

Also given mgf of random variable\[\psi \left( {\rm{t}} \right) = {{\rm{e}}^{{t^2}}}\,{\rm{for}} - \infty < {\rm{t}} < \infty \]

By comparing the given mgf with the mgf of a normal distribution:

We can say that for the given mfg\(\mu = 0\)and\({\sigma ^2} = 2\)

Hence, the random variable follows normal distribution with \(\mu = 0\) and \({\sigma ^2} = 2\)