Mixing
Very basic cumulative Distribution function 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}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
But when we will calculate a probability density function we get :
for $t in [0,2)$ we have $frac{1}{5}$ and for $t notin [0,2)$ we have 0. So it seems that $$int_{-infty}^{infty} ! ϱ_y , mathrm{d}x.=frac{2}{5}$$.
Where is mistake in my justification ?.
probability probability-theory
asked Jan 31 at 22:11
LucianLucian
396
$endgroup$
add a comment |
$begingroup$
Let X ~ U[0,5]. Find cumulative distribution function of Y=min(2,x)
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
for t<0 we have P(Y $le$ t)=0
for t $in$ [0,2) we have P(Y ≤ t)= 1-P(2>x and x>t) = 1-P(x>t)=P(x$le$t)=$frac{1}{5}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
But when we will calculate a probability density function we get :
for $t in [0,2)$ we have $frac{1}{5}$ and for $t notin [0,2)$ we have 0. So it seems that $$int_{-infty}^{infty} ! ϱ_y , mathrm{d}x.=frac{2}{5}$$.
Where is mistake in my justification ?.
probability probability-theory
asked Jan 31 at 22:11
LucianLucian
396
$endgroup$
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41
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}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
But when we will calculate a probability density function we get :
for $t in [0,2)$ we have $frac{1}{5}$ and for $t notin [0,2)$ we have 0. So it seems that $$int_{-infty}^{infty} ! ϱ_y , mathrm{d}x.=frac{2}{5}$$.
Where is mistake in my justification ?.
probability probability-theory
asked Jan 31 at 22:11
LucianLucian
396
$endgroup$
Let X ~ U[0,5]. Find cumulative distribution function of Y=min(2,x)
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
for t<0 we have P(Y $le$ t)=0
for t $in$ [0,2) we have P(Y ≤ t)= 1-P(2>x and x>t) = 1-P(x>t)=P(x$le$t)=$frac{1}{5}$$int_0^t ! 1 , mathrm{d}x.$=$frac{t}{5}$
for t$in$[2,+${displaystyle infty}$) we have P(Y ≤ t)=1
But when we will calculate a probability density function we get :
for $t in [0,2)$ we have $frac{1}{5}$ and for $t notin [0,2)$ we have 0. So it seems that $$int_{-infty}^{infty} ! ϱ_y , mathrm{d}x.=frac{2}{5}$$.
Where is mistake in my justification ?.
probability probability-theory
probability probability-theory
asked Jan 31 at 22:11
LucianLucian
396
asked Jan 31 at 22:11
LucianLucian
396
asked Jan 31 at 22:11
LucianLucian
396
asked Jan 31 at 22:11
LucianLucian
396
asked Jan 31 at 22:11
LucianLucian
396
396
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41
add a comment |
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
No.
$$begin{align}mathsf P(Yleq t) &= mathsf P(Xleq tcap 2leq t) qquad startext{ note: }2leq ttext{, not }x\[1ex] &= mathsf P(Xleq t)cdotmathbf 1_{2leq t}\[1ex]&=begin{cases}0 &:& t<2\2/5 &:& t=2 &startext{ note: this is a point with a probability mass.}\t/5 &:& 2< t < 5\ 1 & :& 5leq tend{cases}end{align}$$
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
No.
$$begin{align}mathsf P(Yleq t) &= mathsf P(Xleq tcap 2leq t) qquad startext{ note: }2leq ttext{, not }x\[1ex] &= mathsf P(Xleq t)cdotmathbf 1_{2leq t}\[1ex]&=begin{cases}0 &:& t<2\2/5 &:& t=2 &startext{ note: this is a point with a probability mass.}\t/5 &:& 2< t < 5\ 1 & :& 5leq tend{cases}end{align}$$
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
$begingroup$
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
No.
$$begin{align}mathsf P(Yleq t) &= mathsf P(Xleq tcap 2leq t) qquad startext{ note: }2leq ttext{, not }x\[1ex] &= mathsf P(Xleq t)cdotmathbf 1_{2leq t}\[1ex]&=begin{cases}0 &:& t<2\2/5 &:& t=2 &startext{ note: this is a point with a probability mass.}\t/5 &:& 2< t < 5\ 1 & :& 5leq tend{cases}end{align}$$
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
$begingroup$
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
No.
$$begin{align}mathsf P(Yleq t) &= mathsf P(Xleq tcap 2leq t) qquad startext{ note: }2leq ttext{, not }x\[1ex] &= mathsf P(Xleq t)cdotmathbf 1_{2leq t}\[1ex]&=begin{cases}0 &:& t<2\2/5 &:& t=2 &startext{ note: this is a point with a probability mass.}\t/5 &:& 2< t < 5\ 1 & :& 5leq tend{cases}end{align}$$
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
$endgroup$
P(Y $le$ t) = P (min(2,x) $le$ t) = 1 - P (min(2,x)>t) = 1-P(2>x and x>t)
No.
$$begin{align}mathsf P(Yleq t) &= mathsf P(Xleq tcap 2leq t) qquad startext{ note: }2leq ttext{, not }x\[1ex] &= mathsf P(Xleq t)cdotmathbf 1_{2leq t}\[1ex]&=begin{cases}0 &:& t<2\2/5 &:& t=2 &startext{ note: this is a point with a probability mass.}\t/5 &:& 2< t < 5\ 1 & :& 5leq tend{cases}end{align}$$
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
answered Jan 31 at 22:49
Graham KempGraham Kemp
87.8k43578
87.8k43578
add a comment |
add a comment |
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
$endgroup$
add a comment |
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
$endgroup$
add a comment |
$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
$endgroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
answered Jan 31 at 22:42
herb steinbergherb steinberg
3,1432311
3,1432311
add a comment |
add a comment |
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$begingroup$
The probability density has a delta function of value $frac{3}{5}$ at $t=2$.
$endgroup$
– herb steinberg
Jan 31 at 22:41