Mixing
$X sim Exp(frac{1}{3}) $ If $Y= max(X,2)$ Find $E(Y)$
$begingroup$
$X sim Exp(frac{1}{3}) $
If $Y= max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
begin{cases}
2, & text{if $0<xle2$ } \[2ex]
X, & text{if $2<x<infty$ }
end{cases}$
$E(Y)=E(Y|0<x<2)P(0<xle2)+E(Y|2<x<infty)P(2<x<infty)$
$E(Y)=2(1-e^{frac{2}{3}})+e^{frac{-2}{3}}int dfrac{x}{3}e^{-frac{x}{3}} dx=2(1-e^{frac{2}{3}})+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2e^{frac{-2}{3}}+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(6e^{-frac{2}{3}}bigg)+bigg(9e^{-frac{2}{3}}bigg)bigg)=2-2e^{-frac{2}{3}}+5e^{-frac{4}{3}}$
Second way
$E(Y)=int max(X,2)f(x)dx=int_0^{2} 2f(x)dx+int _{2}^{infty}xf(x)dx=2(1-e^{frac{-2}{3}})+bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2^{-frac{2}{3}}+5e^{-frac{2}{3}}=2+3e^{-frac{2}{3}}$
Now I am not sure which of them is correct and why can anyone tell me ?
probability statistics
asked Jan 28 at 13:07
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
$X sim Exp(frac{1}{3}) $
If $Y= max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
begin{cases}
2, & text{if $0<xle2$ } \[2ex]
X, & text{if $2<x<infty$ }
end{cases}$
$E(Y)=E(Y|0<x<2)P(0<xle2)+E(Y|2<x<infty)P(2<x<infty)$
$E(Y)=2(1-e^{frac{2}{3}})+e^{frac{-2}{3}}int dfrac{x}{3}e^{-frac{x}{3}} dx=2(1-e^{frac{2}{3}})+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2e^{frac{-2}{3}}+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(6e^{-frac{2}{3}}bigg)+bigg(9e^{-frac{2}{3}}bigg)bigg)=2-2e^{-frac{2}{3}}+5e^{-frac{4}{3}}$
Second way
$E(Y)=int max(X,2)f(x)dx=int_0^{2} 2f(x)dx+int _{2}^{infty}xf(x)dx=2(1-e^{frac{-2}{3}})+bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2^{-frac{2}{3}}+5e^{-frac{2}{3}}=2+3e^{-frac{2}{3}}$
Now I am not sure which of them is correct and why can anyone tell me ?
probability statistics
asked Jan 28 at 13:07
Daman deepDaman deep
756419
$endgroup$
1
$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40
add a comment |
$begingroup$
$X sim Exp(frac{1}{3}) $
If $Y= max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
begin{cases}
2, & text{if $0<xle2$ } \[2ex]
X, & text{if $2<x<infty$ }
end{cases}$
$E(Y)=E(Y|0<x<2)P(0<xle2)+E(Y|2<x<infty)P(2<x<infty)$
$E(Y)=2(1-e^{frac{2}{3}})+e^{frac{-2}{3}}int dfrac{x}{3}e^{-frac{x}{3}} dx=2(1-e^{frac{2}{3}})+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2e^{frac{-2}{3}}+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(6e^{-frac{2}{3}}bigg)+bigg(9e^{-frac{2}{3}}bigg)bigg)=2-2e^{-frac{2}{3}}+5e^{-frac{4}{3}}$
Second way
$E(Y)=int max(X,2)f(x)dx=int_0^{2} 2f(x)dx+int _{2}^{infty}xf(x)dx=2(1-e^{frac{-2}{3}})+bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2^{-frac{2}{3}}+5e^{-frac{2}{3}}=2+3e^{-frac{2}{3}}$
Now I am not sure which of them is correct and why can anyone tell me ?
probability statistics
asked Jan 28 at 13:07
Daman deepDaman deep
756419
$endgroup$
$X sim Exp(frac{1}{3}) $
If $Y= max(X,2)$ Find $E(Y)$
I did in two ways can anyone point out my mistake ?
First one
$f(y)=
begin{cases}
2, & text{if $0<xle2$ } \[2ex]
X, & text{if $2<x<infty$ }
end{cases}$
$E(Y)=E(Y|0<x<2)P(0<xle2)+E(Y|2<x<infty)P(2<x<infty)$
$E(Y)=2(1-e^{frac{2}{3}})+e^{frac{-2}{3}}int dfrac{x}{3}e^{-frac{x}{3}} dx=2(1-e^{frac{2}{3}})+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2e^{frac{-2}{3}}+dfrac{e^{frac{-2}{3}}}{3}bigg(bigg(6e^{-frac{2}{3}}bigg)+bigg(9e^{-frac{2}{3}}bigg)bigg)=2-2e^{-frac{2}{3}}+5e^{-frac{4}{3}}$
Second way
$E(Y)=int max(X,2)f(x)dx=int_0^{2} 2f(x)dx+int _{2}^{infty}xf(x)dx=2(1-e^{frac{-2}{3}})+bigg(bigg(-3xe^{-frac{x}{3}}bigg)_{2}^{infty}-bigg(9e^{-frac{x}{3}}bigg)_{2}^{infty}bigg)=2-2^{-frac{2}{3}}+5e^{-frac{2}{3}}=2+3e^{-frac{2}{3}}$
Now I am not sure which of them is correct and why can anyone tell me ?
probability statistics
probability statistics
asked Jan 28 at 13:07
Daman deepDaman deep
756419
asked Jan 28 at 13:07
Daman deepDaman deep
756419
edited Jan 28 at 13:40
Daman deep
asked Jan 28 at 13:07
Daman deepDaman deep
756419
asked Jan 28 at 13:07
Daman deepDaman deep
756419
asked Jan 28 at 13:07
Daman deepDaman deep
756419
756419
1
$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40
add a comment |
1
$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40
1
1
$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In the first method, you are using the law of total expectation, namely $E(Y)=E,[E(Ymid X)]$.
Note that
begin{align}
E(Y)&=E(Ymid Xle 2)P(Xle 2)+E(Ymid X>2)P(X>2)
\&=E(2mid Xle 2)P(Xle 2)+E(Xmid X>2)P(X>2)
\&=2P(Xle 2)+E(Xmathbf1_{X>2})tag{1}
end{align}
In the second approach, you are using this theorem to calculate
expected value of any function of $X$ directly. I would suggest using this method here.
begin{align}
E(max(X,2))&=frac{1}{3}int max(x,2)e^{-x/3}mathbf1_{x>0},dx
\&=frac{2}{3}int_0^2 e^{-x/3},dx+frac{1}{3}int_2^infty xe^{-x/3},dxtag{2}
end{align}
$(1)$ and $(2)$ are saying the same thing of course.
Integrating by parts the second integral, indeed we get $E(Y)=2+3e^{-2/3}$
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
$endgroup$
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
|
show 5 more comments
$begingroup$
This is how I would proceed (similar to your second method):
$E(Y)=int_0^22timesfrac{1}{3}e^{frac{-x}{3}}dx + int_2^infty xtimesfrac{1}{3}e^{frac{-x}{3}}dx$
$=2(1-e^{frac{-2}{3}}) +3int_frac{2}{3}^infty ttimes e^{-t}dt$
$=2(1-e^{frac{-2}{3}}) +3Gamma(2,frac{2}{3})$
$=2+3e^{frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the first method, you are using the law of total expectation, namely $E(Y)=E,[E(Ymid X)]$.
Note that
begin{align}
E(Y)&=E(Ymid Xle 2)P(Xle 2)+E(Ymid X>2)P(X>2)
\&=E(2mid Xle 2)P(Xle 2)+E(Xmid X>2)P(X>2)
\&=2P(Xle 2)+E(Xmathbf1_{X>2})tag{1}
end{align}
In the second approach, you are using this theorem to calculate
expected value of any function of $X$ directly. I would suggest using this method here.
begin{align}
E(max(X,2))&=frac{1}{3}int max(x,2)e^{-x/3}mathbf1_{x>0},dx
\&=frac{2}{3}int_0^2 e^{-x/3},dx+frac{1}{3}int_2^infty xe^{-x/3},dxtag{2}
end{align}
$(1)$ and $(2)$ are saying the same thing of course.
Integrating by parts the second integral, indeed we get $E(Y)=2+3e^{-2/3}$
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
$endgroup$
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
|
show 5 more comments
$begingroup$
In the first method, you are using the law of total expectation, namely $E(Y)=E,[E(Ymid X)]$.
Note that
begin{align}
E(Y)&=E(Ymid Xle 2)P(Xle 2)+E(Ymid X>2)P(X>2)
\&=E(2mid Xle 2)P(Xle 2)+E(Xmid X>2)P(X>2)
\&=2P(Xle 2)+E(Xmathbf1_{X>2})tag{1}
end{align}
In the second approach, you are using this theorem to calculate
expected value of any function of $X$ directly. I would suggest using this method here.
begin{align}
E(max(X,2))&=frac{1}{3}int max(x,2)e^{-x/3}mathbf1_{x>0},dx
\&=frac{2}{3}int_0^2 e^{-x/3},dx+frac{1}{3}int_2^infty xe^{-x/3},dxtag{2}
end{align}
$(1)$ and $(2)$ are saying the same thing of course.
Integrating by parts the second integral, indeed we get $E(Y)=2+3e^{-2/3}$
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
$endgroup$
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
|
show 5 more comments
$begingroup$
In the first method, you are using the law of total expectation, namely $E(Y)=E,[E(Ymid X)]$.
Note that
begin{align}
E(Y)&=E(Ymid Xle 2)P(Xle 2)+E(Ymid X>2)P(X>2)
\&=E(2mid Xle 2)P(Xle 2)+E(Xmid X>2)P(X>2)
\&=2P(Xle 2)+E(Xmathbf1_{X>2})tag{1}
end{align}
In the second approach, you are using this theorem to calculate
expected value of any function of $X$ directly. I would suggest using this method here.
begin{align}
E(max(X,2))&=frac{1}{3}int max(x,2)e^{-x/3}mathbf1_{x>0},dx
\&=frac{2}{3}int_0^2 e^{-x/3},dx+frac{1}{3}int_2^infty xe^{-x/3},dxtag{2}
end{align}
$(1)$ and $(2)$ are saying the same thing of course.
Integrating by parts the second integral, indeed we get $E(Y)=2+3e^{-2/3}$
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
$endgroup$
In the first method, you are using the law of total expectation, namely $E(Y)=E,[E(Ymid X)]$.
Note that
begin{align}
E(Y)&=E(Ymid Xle 2)P(Xle 2)+E(Ymid X>2)P(X>2)
\&=E(2mid Xle 2)P(Xle 2)+E(Xmid X>2)P(X>2)
\&=2P(Xle 2)+E(Xmathbf1_{X>2})tag{1}
end{align}
In the second approach, you are using this theorem to calculate
expected value of any function of $X$ directly. I would suggest using this method here.
begin{align}
E(max(X,2))&=frac{1}{3}int max(x,2)e^{-x/3}mathbf1_{x>0},dx
\&=frac{2}{3}int_0^2 e^{-x/3},dx+frac{1}{3}int_2^infty xe^{-x/3},dxtag{2}
end{align}
$(1)$ and $(2)$ are saying the same thing of course.
Integrating by parts the second integral, indeed we get $E(Y)=2+3e^{-2/3}$
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
edited Jan 28 at 14:24
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
answered Jan 28 at 14:09
StubbornAtomStubbornAtom
6,30831440
6,30831440
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
|
show 5 more comments
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
I did solve it by parts. Is it wrong ?
$endgroup$
– Daman deep
Jan 28 at 14:13
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
@Damandeep Your final answer is correct.
$endgroup$
– StubbornAtom
Jan 28 at 14:16
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
$begingroup$
Can you tell me in first method what went wrong? I am not able to find mistake. I solved it two three times. There is something wrong .
$endgroup$
– Daman deep
Jan 28 at 14:18
1
1
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
$begingroup$
@Damandeep If you want, in the first method you can use the memoryless property of exponential distribution to say that $E(Xmid X>2)=2+E(X)=2+3=5$. That gives you the answer quickly in fact.
$endgroup$
– StubbornAtom
Jan 28 at 14:34
1
1
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
$begingroup$
oh yes, I calculated it carelessly didn't apply the proper definition of conditional expectation. Thanks, man. That was it!
$endgroup$
– Daman deep
Jan 28 at 15:14
|
show 5 more comments
$begingroup$
This is how I would proceed (similar to your second method):
$E(Y)=int_0^22timesfrac{1}{3}e^{frac{-x}{3}}dx + int_2^infty xtimesfrac{1}{3}e^{frac{-x}{3}}dx$
$=2(1-e^{frac{-2}{3}}) +3int_frac{2}{3}^infty ttimes e^{-t}dt$
$=2(1-e^{frac{-2}{3}}) +3Gamma(2,frac{2}{3})$
$=2+3e^{frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
$endgroup$
add a comment |
$begingroup$
This is how I would proceed (similar to your second method):
$E(Y)=int_0^22timesfrac{1}{3}e^{frac{-x}{3}}dx + int_2^infty xtimesfrac{1}{3}e^{frac{-x}{3}}dx$
$=2(1-e^{frac{-2}{3}}) +3int_frac{2}{3}^infty ttimes e^{-t}dt$
$=2(1-e^{frac{-2}{3}}) +3Gamma(2,frac{2}{3})$
$=2+3e^{frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
$endgroup$
add a comment |
$begingroup$
This is how I would proceed (similar to your second method):
$E(Y)=int_0^22timesfrac{1}{3}e^{frac{-x}{3}}dx + int_2^infty xtimesfrac{1}{3}e^{frac{-x}{3}}dx$
$=2(1-e^{frac{-2}{3}}) +3int_frac{2}{3}^infty ttimes e^{-t}dt$
$=2(1-e^{frac{-2}{3}}) +3Gamma(2,frac{2}{3})$
$=2+3e^{frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
$endgroup$
This is how I would proceed (similar to your second method):
$E(Y)=int_0^22timesfrac{1}{3}e^{frac{-x}{3}}dx + int_2^infty xtimesfrac{1}{3}e^{frac{-x}{3}}dx$
$=2(1-e^{frac{-2}{3}}) +3int_frac{2}{3}^infty ttimes e^{-t}dt$
$=2(1-e^{frac{-2}{3}}) +3Gamma(2,frac{2}{3})$
$=2+3e^{frac{-2}{3}}$
Here is an easy tutorial to solve the incomplete Gamma Integral http://mathworld.wolfram.com/IncompleteGammaFunction.html
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
edited Jan 28 at 14:53
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
answered Jan 28 at 13:52
s0ulr3aper07s0ulr3aper07
580111
580111
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$begingroup$
You meant find $E[Y]$, right?
$endgroup$
– Keen-ameteur
Jan 28 at 13:36
$begingroup$
@keen-ameteur yes I am sorry.
$endgroup$
– Daman deep
Jan 28 at 13:40