Mixing
Finding expectation using cumulative distribution function.
$begingroup$
$ F(x) = begin{cases} 0 & x < -2\ .2& -2 le x< 0 \ .5 & 0
le x <2.2 &\ .6 & 2.2 le x <3 \.9&3 le x<4 \ 1 &
xge4end{cases}$
Find $E(X)$
I am trying to use this formula $E(X)=sum_{x>0}(1-F(x))-sum_{x<0}(F(x))$
$bigg((1-.5)+(1-.6)+(1-.9)+(1-1)bigg)-bigg(.2bigg)=.5+.4+.1-.2=.8$
But this is wrong, answer given is different. Can anyone point out my mistake?
statistics expected-value
asked Jan 26 at 15:26
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
$ F(x) = begin{cases} 0 & x < -2\ .2& -2 le x< 0 \ .5 & 0
le x <2.2 &\ .6 & 2.2 le x <3 \.9&3 le x<4 \ 1 &
xge4end{cases}$
Find $E(X)$
I am trying to use this formula $E(X)=sum_{x>0}(1-F(x))-sum_{x<0}(F(x))$
$bigg((1-.5)+(1-.6)+(1-.9)+(1-1)bigg)-bigg(.2bigg)=.5+.4+.1-.2=.8$
But this is wrong, answer given is different. Can anyone point out my mistake?
statistics expected-value
asked Jan 26 at 15:26
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
$ F(x) = begin{cases} 0 & x < -2\ .2& -2 le x< 0 \ .5 & 0
le x <2.2 &\ .6 & 2.2 le x <3 \.9&3 le x<4 \ 1 &
xge4end{cases}$
Find $E(X)$
I am trying to use this formula $E(X)=sum_{x>0}(1-F(x))-sum_{x<0}(F(x))$
$bigg((1-.5)+(1-.6)+(1-.9)+(1-1)bigg)-bigg(.2bigg)=.5+.4+.1-.2=.8$
But this is wrong, answer given is different. Can anyone point out my mistake?
statistics expected-value
asked Jan 26 at 15:26
Daman deepDaman deep
756419
$endgroup$
$ F(x) = begin{cases} 0 & x < -2\ .2& -2 le x< 0 \ .5 & 0
le x <2.2 &\ .6 & 2.2 le x <3 \.9&3 le x<4 \ 1 &
xge4end{cases}$
Find $E(X)$
I am trying to use this formula $E(X)=sum_{x>0}(1-F(x))-sum_{x<0}(F(x))$
$bigg((1-.5)+(1-.6)+(1-.9)+(1-1)bigg)-bigg(.2bigg)=.5+.4+.1-.2=.8$
But this is wrong, answer given is different. Can anyone point out my mistake?
statistics expected-value
statistics expected-value
asked Jan 26 at 15:26
Daman deepDaman deep
756419
asked Jan 26 at 15:26
Daman deepDaman deep
756419
edited Jan 26 at 15:43
Daman deep
asked Jan 26 at 15:26
Daman deepDaman deep
756419
asked Jan 26 at 15:26
Daman deepDaman deep
756419
asked Jan 26 at 15:26
Daman deepDaman deep
756419
756419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That formula only works when your random variable is confined to the integers, which yours is not (since it can be $2.2$ with positive probability). Instead you can use the formula $E[X]=int_0^infty (1-F(x)) dx - int_{-infty}^0 F(x) dx$. This formula holds for general $X$ provided that the subtraction makes sense (i.e. you don't have $infty-infty$). The formula you wrote is obtained if you split the integrals into the intervals $[n,n+1)$ for integers $n$, provided that $F$ is always constant on such intervals (which again is not the case here).
You can also just write down the PMF and use the usual sum over all possible values. The PMF will be $p(x)=lim_{y to x^+} F(x)-lim_{y to x^-} F(x)$ for all $x$ where $F$ has a jump.
answered Jan 26 at 16:12
IanIan
68.8k25392
$endgroup$
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
add a comment |
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1 Answer
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active
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1 Answer
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active
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$begingroup$
That formula only works when your random variable is confined to the integers, which yours is not (since it can be $2.2$ with positive probability). Instead you can use the formula $E[X]=int_0^infty (1-F(x)) dx - int_{-infty}^0 F(x) dx$. This formula holds for general $X$ provided that the subtraction makes sense (i.e. you don't have $infty-infty$). The formula you wrote is obtained if you split the integrals into the intervals $[n,n+1)$ for integers $n$, provided that $F$ is always constant on such intervals (which again is not the case here).
You can also just write down the PMF and use the usual sum over all possible values. The PMF will be $p(x)=lim_{y to x^+} F(x)-lim_{y to x^-} F(x)$ for all $x$ where $F$ has a jump.
answered Jan 26 at 16:12
IanIan
68.8k25392
$endgroup$
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
add a comment |
$begingroup$
That formula only works when your random variable is confined to the integers, which yours is not (since it can be $2.2$ with positive probability). Instead you can use the formula $E[X]=int_0^infty (1-F(x)) dx - int_{-infty}^0 F(x) dx$. This formula holds for general $X$ provided that the subtraction makes sense (i.e. you don't have $infty-infty$). The formula you wrote is obtained if you split the integrals into the intervals $[n,n+1)$ for integers $n$, provided that $F$ is always constant on such intervals (which again is not the case here).
You can also just write down the PMF and use the usual sum over all possible values. The PMF will be $p(x)=lim_{y to x^+} F(x)-lim_{y to x^-} F(x)$ for all $x$ where $F$ has a jump.
answered Jan 26 at 16:12
IanIan
68.8k25392
$endgroup$
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
add a comment |
$begingroup$
That formula only works when your random variable is confined to the integers, which yours is not (since it can be $2.2$ with positive probability). Instead you can use the formula $E[X]=int_0^infty (1-F(x)) dx - int_{-infty}^0 F(x) dx$. This formula holds for general $X$ provided that the subtraction makes sense (i.e. you don't have $infty-infty$). The formula you wrote is obtained if you split the integrals into the intervals $[n,n+1)$ for integers $n$, provided that $F$ is always constant on such intervals (which again is not the case here).
You can also just write down the PMF and use the usual sum over all possible values. The PMF will be $p(x)=lim_{y to x^+} F(x)-lim_{y to x^-} F(x)$ for all $x$ where $F$ has a jump.
answered Jan 26 at 16:12
IanIan
68.8k25392
$endgroup$
That formula only works when your random variable is confined to the integers, which yours is not (since it can be $2.2$ with positive probability). Instead you can use the formula $E[X]=int_0^infty (1-F(x)) dx - int_{-infty}^0 F(x) dx$. This formula holds for general $X$ provided that the subtraction makes sense (i.e. you don't have $infty-infty$). The formula you wrote is obtained if you split the integrals into the intervals $[n,n+1)$ for integers $n$, provided that $F$ is always constant on such intervals (which again is not the case here).
You can also just write down the PMF and use the usual sum over all possible values. The PMF will be $p(x)=lim_{y to x^+} F(x)-lim_{y to x^-} F(x)$ for all $x$ where $F$ has a jump.
answered Jan 26 at 16:12
IanIan
68.8k25392
edited Jan 26 at 16:21
answered Jan 26 at 16:12
IanIan
68.8k25392
answered Jan 26 at 16:12
IanIan
68.8k25392
answered Jan 26 at 16:12
IanIan
68.8k25392
68.8k25392
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
add a comment |
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Could you let us know the PMF method? I will delete mine.
$endgroup$
– Satish Ramanathan
Jan 26 at 16:19
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
Converting into PMF then the calculating $E(X)$ method I know I wanted to know what is the issue in the formula I wrote. Thanks for clearing my doubt.
$endgroup$
– Daman deep
Jan 26 at 16:20
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
$begingroup$
I just posted a question regarding same concept can you take a look ?
$endgroup$
– Daman deep
Feb 8 at 4:28
add a comment |
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