Question

If x is uniformly distributed over (0,15) .The probability that 5

Answer 1

Note that 33 is not the density of XX. As XX is uniformly distributed, XX has a constant density fX(x)=cfX(x)=c, for some c∈Rc∈R. As we must have

1=P(−5≤X≤5)=∫5−5fX(x)dx=∫5−5cdx=10c

1=P(−5≤X≤5)=∫−55fX(x)dx=∫−55cdx=10c

we have fX(x)=110fX(x)=110. Now continue as you did, we have

P(|X|>3)=∫−3−5110dx+∫53110dx=25

P(|X|>3)=∫−5−3110dx+∫35110dx=25

For the density of |X||X|, let a∈(0,5)a∈(0,5), we have

P(|X|<a)=∫a−afX(x)dx=a5

P(|X|<a)=∫−aafX(x)dx=a5

Taking derivatives, as you did correctly, we have

f|X|(a)=15,a∈(0,5)

f|X|(a)=15,a∈(0,5)

Hence, |X||X| is uniformly distributed on (0,5)(0,5).