Question

tso From aeda : Des H. (o) exist?

Answer 1

(a)

Since we see a summation in the definition of MGF of X we suspect that The distribution of X is discretre.

Now, the mgf of X is given by the formula :

$\small M_X(t)= E(e^{tX})=\sum_{x} P(X=x)*e^{tx}$

Comparing the above expression with the given expression $\small \sum_{x=1}^\infty \frac{6}{\pi^2}*\frac{1}{x^2}*e^{tx} \; \; ; t \leq 0$

we observe that the support of X is {1,2,3,.....}

and $\small P(X=x)=\frac{6}{\pi^2}*\frac{1}{x^2} \; \; \; ;x=1,2,3,...$ , which is the probability mass function of X.

(b)

$\small E(X) = \sum_{x=1}^{\infty} x*P(X=x) = \sum_{x=1}^{\infty} x*\frac{6}{\pi^2}*\frac{1}{x^2} = \frac{6}{\pi^2} \sum_{x=1}^{\infty} \frac{1}{x}$

Now, we know that the series $\small \sum_{x=1}^{\infty} \frac{1}{x}$ diverges. Thus, E(X) does not exist.

(c)

To check whether $\small M_X'(0)$ exists or not, we calculate the left hand limit and right hand limit of the derivative of MGF at zero.

RHL : $\small \lim_{t \to 0^+} \frac{M_X(t)-M_X(0)}{t-0}$

Now, it is the property of MGF's that $\small M_X(0) = 1$

Thus, RHL : $\small \lim_{t \to 0^+} \frac{M_X(t)-1}{t}$

Now, $\small M_X(t) = +\infty$ when t > 0.

Thus, RHL = $\small +\infty$

Also, $\small \lim_{t \to 0^-} \frac{M_X(0)-M_X(t)}{0-t} = \lim_{t \to 0^-} \frac{1- \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}*e^{tn}) }{-t}$

$\small Now, \; \; \lim_{t \to 0^-} (1- \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}*e^{tn}) ) = 1- \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}*e^{0}) \newline = 1- \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}) = 1- \frac{6}{\pi^2}*\frac{\pi^2}{6} = 1-1 = 0$

Thus, we get $\small \frac{0}{0}$ indeterminate form and we apply L-Hopital's Rule.

$\small LHL = \lim_{t \to 0^-} \frac{ \frac{\mathrm{d} }{\mathrm{d} t}(1- \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}*e^{tn})) }{\frac{\mathrm{d} }{\mathrm{d} t}(-t)} = \lim_{t \to 0^-} \frac{-\frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n^2}*n*e^{tn})}{-1} \newline = \lim_{t \to 0^-} \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n}*e^{tn}) = \frac{6}{\pi^2} *\sum_{n=1}^\infty (\frac{1}{n}) = +\infty$

Thus, $\small M_X'(0)$ exists and is equal to $\small +\infty$