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Exer. 2. Prove THAT $M_g(t) = \int_{-\infty}^{+\infty} e^{tx} \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} dx =$ $= e^{\frac{t^2}{2}} !!! Hint: $e^{tx} + e^{-\frac{x^2}{2}} = \frac{1}{\sqrt{2\pi}} e^{-(x-t)^2 \frac{t^2}{2}} e^{\frac{t^2}{2}}$ So $M_g(t) = e^{\frac{t^2}{2}} \int_{-\infty}^{+\infty} \frac{e^{-\frac{(x-t)^2}{2}}}{\sqrt{2\pi}} dx$ (ave MORE CHANGE OF VARIABLE HERE!) Exer. 3. Prove that k is odd $(e^{\frac{t^2}{2}})|_{t=0} = \begin{cases} 0, & k \frac{(2m)!}{2^m m!}, & k = 2m (kis \text{even}) \end{cases}$ So, $E(\frac{t^2}{2}m) = \frac{(2m)!}{2^m m!}$ Hint: f(x) = $\sum_{k=0}^{\infty} a_k x^k$ $\implies$ TAYLOR'S !!! FORMULA $f^{(k)}(0) = a_k k!$ 13

          Exer. 2.
Prove THAT
$M_g(t) = \int_{-\infty}^{+\infty} e^{tx} \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}} dx =$
$= e^{\frac{t^2}{2}} !!!
Hint: $e^{tx} + e^{-\frac{x^2}{2}} = \frac{1}{\sqrt{2\pi}} e^{-(x-t)^2 \frac{t^2}{2}} e^{\frac{t^2}{2}}$
So $M_g(t) = e^{\frac{t^2}{2}} \int_{-\infty}^{+\infty} \frac{e^{-\frac{(x-t)^2}{2}}}{\sqrt{2\pi}} dx$ (ave
MORE CHANGE OF VARIABLE HERE!)
Exer. 3. Prove that k is odd
$(e^{\frac{t^2}{2}})|_{t=0} = \begin{cases} 0, & k \frac{(2m)!}{2^m m!}, & k = 2m (kis \text{even}) \end{cases}$
So, $E(\frac{t^2}{2}m) = \frac{(2m)!}{2^m m!}$
Hint:
f(x) = $\sum_{k=0}^{\infty} a_k x^k$
$\implies$ TAYLOR'S !!!
FORMULA
$f^{(k)}(0) = a_k k!$ 13

Exer. 2.
Prove THAT
Mg(t) = ∫-∞^+∞ e^tx(e^-(x^2)/(2))/(√(2π)) dx =
= e^(t^2)/(2) !!!
Hint:e^tx + e^-(x^2)/(2) = (1)/(√(2π)) e^-(x-t)^2 (t^2)/(2) e^(t^2)/(2)SoMg(t) = e^(t^2)/(2) ∫-∞^+∞ (e^-((x-t)^2)/(2))/(√(2π)) dx(ave
MORE CHANGE OF VARIABLE HERE!)
Exer. 3. Prove that k is odd(e^(t^2)/(2))|t=0 = 0, k ((2m)!)/(2^m m!), k = 2m (kis even) So,E((t^2)/(2)m) = ((2m)!)/(2^m m!)Hint:
f(x) =∑k=0^∞ ak x^kTAYLOR'S !!!
FORMULAf^(k)(0) = ak k!13

Added by Cheyenne W.

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