Let X ~ U[0,5]. Find cumulative distribution function of Y=min(2,x)

P(Y $\le$ t) = P (min(2,x) $\le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)

for t<0 we have P(Y $\le$ t)=0

for t $\in$ [0,2) we have P(Y ≤ t)= 1-P(2>x and x>t) = 1-P(x>t)=P(x$\le$t)=$\frac{1}{5}$$\int_0^t \! 1 \, \mathrm{d}x.$=$\frac{t}{5}$

for t$\in$[2,+${\displaystyle \infty}$) we have P(Y ≤ t)=1

Where is the mistake ?

  • 2
    $\begingroup$ Why do you think there is a mistake? $\endgroup$ – Henry Jan 25 at 23:23
  • $\begingroup$ Looks good to me. $\endgroup$ – herb steinberg Jan 26 at 0:55
  • $\begingroup$ We can calculate what's the probability density function of Y and we have : For t$\notin$[0,2) we have 0 For t$\in$[0,2) we have 1/5. Then $\int_{-{\displaystyle \infty}}^{{\displaystyle \infty}}$ ϱ$^Y$=$\frac{2}{5}$ $\endgroup$ – Lucian Jan 26 at 8:21

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