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You decide to simulate a random variable. You generate a value on $U(0,1)$ of $.618$. What is your simulated value of $X$ if:

a) $X$ is $U(0,10)$?

b) $X$ follows $F(x) = 1 - e^{-2x}$?

c) $X$ is $N(0,1)$?

Attempted Solutions:

a) I am assuming $U(0,1)$ means every possible outcome in the event space is equally likely to occur. I am not sure then what the simulated value would be for $X$ but I would guess $6.18$?

b) I am not sure where to start. I recall seeing that $f(x) = F'(x)$ so $f(x) = 2e^{-2x}$

c) Apparently, $N(0,1)$ means $N(\mu, \sigma^2)$ and wiki says the curve in this special case where $\mu = 0$ and $\sigma^2 = 1$ maxes out at $1\over{\sqrt{2\pi}}$ so I suppose that would be the answer

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You are supposed to find the value $X$ for each distribution that has $0.618$ of the area to the left of it. This would be represented as $CDF(X)=0.618$ or $\int_{-\infty}^X pdf(x)dx = 0.618$ In a, you are correct it is $6.18$. For b, you are given the CDF, so solve the equation. For c, you need to use the z-score table for the normal distribution.

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    For (b) do I set pdf = $2e^{2x}$ and look for the upper limit that makes it equal to .618? – 2017-02-07
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    That would work, but you have the CDF so can use it. You just want $0.618=1-e^{-2X}$. – 2017-02-07
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    For (c) I got .3 because the z score associated with .618 is .3 and a standard deviation is just 1. Is that right? – 2017-02-07
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    Yes, that is right for c – 2017-02-07
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    Cool, and for (b) I obtained .4812 – 2017-02-07
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    Appreciate the help by the way, thank you – 2017-02-07