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| Boilerpipe Text | We divide each side of the inequality by $\sigma/\sqrt{2}$ , which is how we get from the first circled step to the next. If you're asking why we're dividing, it's because we want to make the random variable on the left-hand side of the inequality a standard normal random variable. This way, we can use the normal cumulative distribution function to finish our calculation. |
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# [An example with Brownian Motion](https://math.stackexchange.com/questions/3604197/an-example-with-brownian-motion)
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\$\\begingroup\$

I wonder why here in the solution above, Why P{Y(1/2)\> - segema} becomes P{ Y(1/2)/segema/ square root of2}
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\$\\begingroup\$ Hi cynthiaaaaii, thanks for posting the question. Can you copy the text in your image to the description because this will make the problem easy to find later. \$\\endgroup\$
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\$\\begingroup\$
We divide each side of the inequality by \$\\sigma/\\sqrt{2}\$, which is how we get from the first circled step to the next.
If you're asking why we're dividing, it's because we want to make the random variable on the left-hand side of the inequality a standard normal random variable. This way, we can use the normal cumulative distribution function to finish our calculation.
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answered Apr 1, 2020 at 1:24
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