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$X\sim$ Uniform$(0, 1)$ and $Y\sim$ Uniform$(0, 1)$ compute $P(X+Y\ge0.5)$

I tried to mark $Z=X+Y$ therefore $Z\sim N(0,2)$

$$P(X+Y\ge0.5) = 1- P(X+Y<0.5) = 1- P(Z<0.5) = f_z(0.5)\\ = \frac1{(2\sqrt{\pi}) } \times \exp(-1/4\times\sqrt2)$$

not sure if it is correct.

Thanks

Gary
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greenity
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1 Answers1

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Disclaimer: In this homework-type question, $X$ and $Y$ are probably assumed to be statistically independent. My answer assumes this, although you didn't state it explicitly.

State the joint PDF of $X$ and $Y$ (hint: it's constant $1$ in a square and $0$ otherwise) and calculate $1 - P(X + Y < 0.5)$ instead. Identify the area of duples in the square $(x,y) \in [0,1]^2$ that fulfill $x + y < 0.5$ and calculate the probability by calculating the area of the resulting trianlge (easy).

P.S.: As mentioned by Henry, your assumption $N(0,2)$ is bogus indeed.

GDumphart
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  • You cannot find the joint distribution function of $(X,Y)$ or the distribution function of $X$ and $Y$ from the individual distribution functions without further assumptions. – Kavi Rama Murthy Jan 09 '22 at 11:29
  • True, but in this homework type question, $X$ and $Y$ are almost certainly intended to be iid. Even if not stated explicitly. – GDumphart Jan 09 '22 at 11:31
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    Added a disclaimer to my answer. – GDumphart Jan 09 '22 at 11:33
  • thanks for the explanation but I am stack in the computation itself. by the definition of the pdf . I can agree that z!~N(0,2) but it is not helpful even though – greenity Jan 09 '22 at 13:11