# cumulative distribution function. Basic problem

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 ?

• Why do you think there is a mistake? – Henry Jan 25 at 23:23
• Looks good to me. – herb steinberg Jan 26 at 0:55
• 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$^Y$=$\frac{2}{5}$ – Lucian Jan 26 at 8:21