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$\displaystyle\int_{0}^{1} e^{as}dB_{s}\leq\int_{0}^{2} e^{as}dB_{s}$, is this true?

In my opinion,

$\displaystyle\int_{0}^{1} e^{as}dB_{s}=\int_{0}^{1} e^{as}dB_{s}+\int_{1}^{2} e^{as}dB_{s}$ and $\int_{1}^{2} e^{as}dB_{s}\sim N(0,\frac{e^{2as}-e^{as}}{as})$.

where $B_{t}$ is a Brownian motion. Thanks for all your help?

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    so your opinion is that it's false? (also what is $\sigma^2$?... you can compute it, or does it matter? ) – 2017-01-09
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    @spaceisdarkgreen Thanks. Please see the edited question again. – 2017-01-09
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    You are adding an independent centered normal variable, which can be both negative and positive. – 2017-01-09
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    That distribution you wrote down looks about right (I think you might have missed a factor of two in the variance, but I don't know that it's important.). So what's your conclusion? Why were you wondering if it was true in the first place? – 2017-01-09
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    @spaceisdarkgreen Because I want to use the well known inequlity ( Chebyshev's sum inequality) $\sum_{k=1}^{n}a_{k}b_{k}\leq \frac{1}{n}(\sum_{k=1}^{n}a_{k})(\sum_{k=1}^{n}b_{k})$. So I should testify the monotonicity of $\int_{0}^{1} e^{as}dB_{s},\int_{0}^{3} e^{as}dB_{s},\int_{0}^{2} e^{as}dB_{s}$ and so on. – 2017-01-09
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    But your argument shows the inequality is not true. The difference is normally distributed, thus can be positive or negative. – 2017-01-09
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    @spaceisdarkgreen Thanks for your kindly help. – 2017-01-09

1 Answers 1

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$\displaystyle P\bigg(\int_0^1e^{as}dB_s\le\int_0^2e^{as}dB_s\bigg)=P\bigg(\int_1^2e^{as}dB_s\ge 0\bigg)=1/2$, since $\int_1^2e^{as}dB_s\sim N(0,\sigma^2)$ and $\sigma^2>0$.

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    impressive of your answer. – 2017-01-09