exercise:A4a4728515: Difference between revisions
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Let <math>X_i\sim\mathcal{N}(0,1)</math> and <math>X=X_1+\dots+X_d</math>. | |||
<ul style="list-style-type:lower-roman"><li> Use the formula <math>\E(f(X_i))=(2\pi)^{-1/2}\int_{\mathbb{R}}f(x)\exp(-x^2/2)\dd x</math> to show that <math>\E(\exp(tX_i))=(1-2t)^{-d/2}</math> holds for <math>t\in(0,1/2)</math>. | <ul style="list-style-type:lower-roman"><li> Use the formula <math>\E(f(X_i))=(2\pi)^{-1/2}\int_{\mathbb{R}}f(x)\exp(-x^2/2)\dd x</math> to show that <math>\E(\exp(tX_i))=(1-2t)^{-d/2}</math> holds for <math>t\in(0,1/2)</math>. | ||
</li> | </li> | ||
Latest revision as of 02:46, 2 June 2024
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[/math]
Let [math]X_i\sim\mathcal{N}(0,1)[/math] and [math]X=X_1+\dots+X_d[/math].
- Use the formula [math]\E(f(X_i))=(2\pi)^{-1/2}\int_{\mathbb{R}}f(x)\exp(-x^2/2)\dd x[/math] to show that [math]\E(\exp(tX_i))=(1-2t)^{-d/2}[/math] holds for [math]t\in(0,1/2)[/math].
- Derive the estimate [math]\P\bigl[X\geqslant a\bigr] \leqslant\inf_{t\in(0,1/2)}\frac{\exp(-ta)}{(1-2t)^{d/2}}[/math] for [math]a \gt 0[/math].