If $X$ follows $N(0,1)$ what is the mgf of Y?
I think I found that the pdf of Y is $\frac{1}{\sqrt{2y\pi}}e^{(\sqrt{y}+3)t}e^{\frac{-(\sqrt{y}+3)^2}{2}}$
Then I think you're supposed to integrate but the integral doesn't work out well.
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1What have you tried? Please edit your work into the question body. – Varun Vejalla Sep 27 '20 at 00:56
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what is the definition of mgf? – Sep 27 '20 at 01:56
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Using the definition, $$\begin{aligned}M_Y(t):=\mathbb{E}\left[e^{Yt}\right]&=\mathbb{E}\left[e^{(X-3)^2t}\right]=\frac1{\sqrt{2\pi}}\int_{\mathbb{R}} e^{t(x-3)^2}e^{-\frac12x^2}dx\\ &=\frac{e^{9t+\frac{18 t^2}{1-2 t}}}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-\frac12\left[(1-2t)x^2+12tx+\frac{36 t^2}{1-2 t}\right]}dx\\ &=\frac{e^{\frac{9 t}{1-2 t}}}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-\frac12\left[x\sqrt{1-2t}+\frac{6 t}{\sqrt{1-2 t}}\right]^2}dx\\ \end{aligned}$$ now let $u=x\sqrt{1-2t}$, $$\begin{aligned} \frac{e^{\frac{9 t}{1-2 t}}}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-\frac12\left[x\sqrt{1-2t}+\frac{6 t}{\sqrt{1-2 t}}\right]^2}dx&=\frac1{\sqrt{1-2t}}e^{\frac{9 t}{1-2 t}}\int_{\mathbb{R}}\frac1{\sqrt{2\pi}} e^{-\frac12\left[u-\frac{-6 t}{\sqrt{1-2 t}}\right]^2}du\\ &=\frac1{\sqrt{1-2t}}e^{\frac{9 t}{1-2 t}}\cdot1=\frac1{\sqrt{1-2t}}e^{\frac{9 t}{1-2 t}},\,\,t<\frac12. \end{aligned}$$
kiyomi
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