91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

(a) pdf of normal distribution is given as:

\({\rm{f(x;\mu ,\sigma ) = }}\frac{{\rm{1}}}{{{\rm{\sigma }}\sqrt {{\rm{2\pi }}} }}{\rm{ \times }}{{\rm{e}}^{\frac{{{\rm{ - (x - \mu }}{{\rm{)}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}}}\)

For \({\rm{f(x)}}\) to be maximum at\({{\rm{x}}^{\rm{*}}}\), it's derivative wrt \({\rm{x}}\) has to be equal to 0 at \({{\rm{x}}^{\rm{*}}}\)

\(\begin{aligned}\frac{d}{{dx}}f(x)&= \frac{1}{{\sigma \sqrt {2\pi } }} \times {e^{\frac{{ - {{(x - \mu )}^2}}}{{2{\sigma ^2}}}}} \times \frac{{ - 2(x - \mu )}}{{2{\sigma ^2}}}{\left[ {\frac{d}{{dx}}f(x)} \right]_{x = {x^*}}} \\&= \frac{{ - \left( {{x^*} - \mu } \right)}}{{{\sigma ^3}\sqrt {2\pi } }} \times {e^{\frac{{ - {{\left( {{x^*} - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}0 \\&= \frac{{ - \left( {{x^*} - \mu } \right)}}{{{\sigma ^3}\sqrt {2\pi } }} \times {e^{\frac{{ - {{\left( {{x^*} - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}{x^*} \\&= \mu \\\end{aligned} \)

(b) No, the uniform distribution with parameters \({\rm{A}}\) and \({\rm{B}}\) have a single mode. Because it's pdf is constant for all the values in the interval \({\rm{A}}\) to\({\rm{B}}\).

(c) pdf of exponential distribution with parameter \({\rm{\lambda }}\) is given as:

\({\rm{f(x;\lambda ) = \lambda \times }}{{\rm{e}}^{{\rm{ - \lambda x}}}}\)

For \({\rm{f(x)}}\) to be maximum at\({{\rm{x}}^{\rm{*}}}\), it's derivative wrt \({\rm{x}}\) has to be equal to 0 at \({{\rm{x}}^{\rm{*}}}\)

\(\begin{aligned}\frac{d}{{dx}}f(x) &= \lambda \times {e^{ - \lambda x}} \times ( - \lambda ){\left[ {\frac{d}{{dx}}f(x)} \right]_{x = {x^*}}}\\&=- {\lambda ^2} \times {e^{-\lambda {x^*}}}\\ \end{aligned} \)

As we can see that the derivative does not become equal to \({\rm{0}}\) at any value of \({\rm{x}}\). But by observing pdf\({\rm{f(x)}}\), we see that it is maximum at \({\rm{x = 0}}\) and decreases as \({\rm{x}}\) increases. Hence

\({{\rm{x}}^{\rm{*}}}{\rm{ = 0}}\)

A plot of exponential distribution with \({\rm{\lambda = 2}}\) is given in the figure below:

(d) pdf of gamma distribution with parameters \({\rm{\alpha }}\) and\({\rm{\beta }}\) is given as:

\({\rm{f(x;\alpha ,\beta ) = }}\frac{{\rm{1}}}{{{{\rm{\beta }}^{\rm{\alpha }}}{\rm{ \times \Gamma (\alpha )}}}}{\rm{ \times }}{{\rm{x}}^{{\rm{\alpha - 1}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - x/\beta }}}}\)

As given as hint, let us consider a function \({\rm{g(x)}}\) such that:

\(\begin{array}{l}{\rm{g(x) = ln(f(x))}}\\{\rm{g(x) = - ln}}\left( {{{\rm{\beta }}^{\rm{\alpha }}}{\rm{ \times \Gamma (\alpha )}}} \right){\rm{ + (\alpha - 1) \times ln(x) - }}\frac{{\rm{x}}}{{\rm{\beta }}}\end{array}\)

Hence \({\rm{g(x)}}\) will be maximum if \({\rm{f(x)}}\) is maximum. Hence at\({\rm{x = }}{{\rm{x}}^{\rm{*}}}\):

\(\begin{aligned}{\left[ {\frac{d}{{dx}}g(x)} \right]_{x = {x^*}}} &= {\left[ {\frac{d}{{dx}}f(x)} \right]_{x = {x^*}}}\\&= 0{\left[ {\frac{d}{{dx}}g(x)} \right]_{x = {x^*}}} \\&= \frac{{\alpha - 1}}{{{x^*}}} - \frac{1}{\beta }0 \\&= \frac{{\alpha - 1}}{{{x^*}}} - \frac{1}{\beta }{x^*} \\&= \frac{{\alpha - 1}}{\beta } \\\end{aligned} \)

(e) pdf of chi-squared distribution with degree of freedom \({\rm{\nu }}\) is given as:

\({\rm{f(x;\nu ) = }}\frac{{\rm{1}}}{{{{\rm{2}}^{{\rm{\nu /2}}}}{\rm{ \times \Gamma (\nu /2)}}}}{\rm{ \times }}{{\rm{x}}^{{\rm{\nu /2 - 1}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - x/2}}}}\)

Let us consider a function \({\rm{g(x)}}\) such that:

\(\begin{array}{l}{\rm{g(x) = ln(f(x))}}\\{\rm{g(x) = - ln}}\left( {{{\rm{2}}^{{\rm{\nu /2}}}}{\rm{ \times \Gamma (\nu /2)}}} \right){\rm{ + }}\left( {\frac{{\rm{\nu }}}{{\rm{2}}}{\rm{ - 1}}} \right){\rm{ \times ln(x) - }}\frac{{\rm{x}}}{{\rm{2}}}\end{array}\)

Hence \({\rm{g(x)}}\) will be maximum if \({\rm{f(x)}}\) is maximum. Hence at\({\rm{x = }}{{\rm{x}}^{\rm{*}}}\):

\(\begin{aligned}{\left[ {\frac{d}{{dx}}g(x)} \right]_{x = {x^*}}} &= {\left[ {\frac{d}{{dx}}f(x)} \right]_{x = {x^*}}} \\&= 0{\left[ {\frac{d}{{dx}}g(x)} \right]_{x = {x^*}}} \\&= \left( {\frac{\nu }{2} - 1} \right) \times \frac{1}{{{x^*}}} - \frac{1}{2}0 \\&= \left( {\frac{\nu }{2} - 1} \right) \times \frac{1}{{{x^*}}} - \frac{1}{2}{x^*} \\&= \nu - 2 \\\end{aligned} \)