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Let $X$, $Y$ be two independent random variables from $U(0,1)$. Then find $P[Y>(X-1/2)^2]$.

I initially tried drawing the figure but that seemed complicated. I then took expectation on both sides and got $P[E(Y)>V(X)]$. Am I right?

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    $\begingroup$ What did you attempt to draw? The drawing is very simple so the area of integration should be quite simple to figure out once you sketch the x-y plane. $\endgroup$ – StatsStudent Apr 22 '17 at 5:50
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    $\begingroup$ Also, why are you doing anything with expectation? You are simply asked to find the probability, not the expectation. $\endgroup$ – StatsStudent Apr 22 '17 at 5:59
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    $\begingroup$ @Analyst1 Probabilities are expectations (of indicator variables). Often they can readily be found using techniques to find expectations. The mistake made here is to suppose that $E[Y\gt (X-\mu_X)^2] = E[Y \gt E[(X-\mu_X)^2]]$: this is rarely true and there are no generally applicable rules of expectation, probability, or integration that would justify such an equation. $\endgroup$ – whuber Apr 22 '17 at 12:02
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    $\begingroup$ Good point @whuber, but I don't see where the OP was headed with regard to his/her approach by finding an expectation. The most straightforward approach to solving this problem seemed to be simply integrating the area bounded below by 0 and 1 and above the parabola $Y = (X-1/2)^2$. $\endgroup$ – StatsStudent Apr 22 '17 at 15:48
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This ends up being the area above the curve \begin{equation} Y=(X−\frac{1}{2})^2 \end{equation}

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This can be found by integration \begin{equation} P[Y>(X−\frac{1}{2})^2] = \int_{0}^{1}\int_{(X−\frac{1}{2})^2}^{1} 1\times1 \,dydx \end{equation} \begin{equation} = \int_{0}^{1}{1-(X−\frac{1}{2})^2} \,dx \end{equation} \begin{equation} = \Big[X-\frac{1}{3}\times(X−\frac{1}{2})^3\Big]_0^1 \end{equation} \begin{equation} = \Big(1-\frac{1}{3}\times(1−\frac{1}{2})^3\Big) - \Big(0-\frac{1}{3}\times(0−\frac{1}{2})^3\Big) \end{equation} \begin{equation} = \frac{23}{24}-\frac{1}{24} \end{equation} \begin{equation} = \frac{22}{24} \end{equation}

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