Mixing
cumulative distribution function. Basic problem
$begingroup$
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}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
Where is the mistake ?
probability
asked Jan 25 at 23:06
LucianLucian
396
$endgroup$
add a comment |
$begingroup$
Let X ~ U[0,5]. Find cumulative distribution function of Y=min(2,x)
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
for t<0 we have P(Y $le$ t)=0
for t $in$ [0,2) we have P(Y ≤ t)= 1-P(2>x and x>t) = 1-P(x>t)=P(x$le$t)=$frac{1}{5}$$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 ?
probability
asked Jan 25 at 23:06
LucianLucian
396
$endgroup$
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
add a comment |
$begingroup$
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}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
Where is the mistake ?
probability
asked Jan 25 at 23:06
LucianLucian
396
$endgroup$
Let X ~ U[0,5]. Find cumulative distribution function of Y=min(2,x)
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
for t<0 we have P(Y $le$ t)=0
for t $in$ [0,2) we have P(Y ≤ t)= 1-P(2>x and x>t) = 1-P(x>t)=P(x$le$t)=$frac{1}{5}$$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 ?
probability
probability
asked Jan 25 at 23:06
LucianLucian
396
asked Jan 25 at 23:06
LucianLucian
396
edited Jan 26 at 8:21
Lucian
asked Jan 25 at 23:06
LucianLucian
396
asked Jan 25 at 23:06
LucianLucian
396
asked Jan 25 at 23:06
LucianLucian
396
396
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
add a comment |
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
2
2
$begingroup$
Why do you think there is a mistake?
$endgroup$
– Henry
Jan 25 at 23:23
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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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