Calculate the variance of $f_X(x)$
$begingroup$
We have $$fleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{50}left(x-50right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{128}left(x-60right)^2}right)$$
This can be written in a more nicer way:
$$gleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{2}left(frac{x-50}{5}right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{2}left(frac{x-60}{8}right)^2}right)$$
To calculate the expectation, it is equal to $0.8(50) + 0.2(60)=52$, since we see that this is essentially $0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)$
We get confirmation from desmos:
What about variance though?
$var(0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)=0.8^2(25)+0.2^2(64)=18.56$
But also, $v(x)=E(x^2)-(E(x))^2$
Verifying the above on desmos, I get a different answer:
About $48.8$, so I am not sure where I went wrong. I know the integral is entered correctly on desmos, but where have I gone wrong with the variance?
probability probability-distributions normal-distribution variance
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
$endgroup$
add a comment |
$begingroup$
We have $$fleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{50}left(x-50right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{128}left(x-60right)^2}right)$$
This can be written in a more nicer way:
$$gleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{2}left(frac{x-50}{5}right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{2}left(frac{x-60}{8}right)^2}right)$$
To calculate the expectation, it is equal to $0.8(50) + 0.2(60)=52$, since we see that this is essentially $0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)$
We get confirmation from desmos:
What about variance though?
$var(0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)=0.8^2(25)+0.2^2(64)=18.56$
But also, $v(x)=E(x^2)-(E(x))^2$
Verifying the above on desmos, I get a different answer:
About $48.8$, so I am not sure where I went wrong. I know the integral is entered correctly on desmos, but where have I gone wrong with the variance?
probability probability-distributions normal-distribution variance
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
$endgroup$
add a comment |
$begingroup$
We have $$fleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{50}left(x-50right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{128}left(x-60right)^2}right)$$
This can be written in a more nicer way:
$$gleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{2}left(frac{x-50}{5}right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{2}left(frac{x-60}{8}right)^2}right)$$
To calculate the expectation, it is equal to $0.8(50) + 0.2(60)=52$, since we see that this is essentially $0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)$
We get confirmation from desmos:
What about variance though?
$var(0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)=0.8^2(25)+0.2^2(64)=18.56$
But also, $v(x)=E(x^2)-(E(x))^2$
Verifying the above on desmos, I get a different answer:
About $48.8$, so I am not sure where I went wrong. I know the integral is entered correctly on desmos, but where have I gone wrong with the variance?
probability probability-distributions normal-distribution variance
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
$endgroup$
We have $$fleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{50}left(x-50right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{128}left(x-60right)^2}right)$$
This can be written in a more nicer way:
$$gleft(xright)=0.8left(frac{1}{5sqrt{2pi}}e^{-frac{1}{2}left(frac{x-50}{5}right)^2}right)+0.2left(frac{1}{8sqrt{2pi}}e^{-frac{1}{2}left(frac{x-60}{8}right)^2}right)$$
To calculate the expectation, it is equal to $0.8(50) + 0.2(60)=52$, since we see that this is essentially $0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)$
We get confirmation from desmos:
What about variance though?
$var(0.8N(mu=50, sigma^2=25)+0.2N(mu=60, sigma^2=64)=0.8^2(25)+0.2^2(64)=18.56$
But also, $v(x)=E(x^2)-(E(x))^2$
Verifying the above on desmos, I get a different answer:
About $48.8$, so I am not sure where I went wrong. I know the integral is entered correctly on desmos, but where have I gone wrong with the variance?
probability probability-distributions normal-distribution variance
probability probability-distributions normal-distribution variance
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
asked Dec 16 '18 at 22:26
K Split XK Split X
4,30421233
4,30421233
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
$f_X$ is the density of the random variable $UZ_1+(1-U)Z_2$ where $Usim Bernoulli(0.8), Z_1sim N(50,25)$ and $Z_2sim N(60,64)$ and they are mutually independent.
Then $Var(UZ_1+(1-U)Z_2)=E[Var(UZ_1+(1-U)Z_2|U)]+Var[E(UZ_1+(1-U)Z_2|U)].$
$E[Var(UZ_1+(1-U)Z_2|U)]=E[U^2Var(Z_1|U)+(1-U)^2Var(Z_2|U)]=E[U^2]Var(Z_1)+E[(1-U)^2]Var(Z_2)=E[U]times 25+E[1-U]times 64=0.8times 25+0.2times 64=32.8.$
$Var[E(UZ_1+(1-U)Z_2|U)]=Var[UE(Z_1)+(1-U)E(Z_2)]=Var[50U+60(1-U)]=Var[-10U]=100times 0.8times 0.2=16.$
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
$endgroup$
add a comment |
$begingroup$
for a function of the form
$$
f(x) = a frac{1}{sqrt{2pi sigma_1^2}}e^{(x - mu_1)^2/2sigma_1^2} + (1- a)frac{1}{sqrt{2pi sigma_2^2}}e^{(x - mu_2)^2/2sigma_2^2}
$$
you have
$$
int_{-infty}^{+infty} {rm d}x~ f(x) = a + (1 - a) = 1
$$
$$
int_{-infty}^{+infty} {rm d}x~ x f(x) = amu_1 + (1 - a)mu_2 tag{1}
$$
and
$$
int_{-infty}^{+infty} {rm d}x~ x^2 f(x) = a(mu_1^2 + sigma_1^2) + (1 - a)(mu_2^2 + sigma_2^2) tag{2}
$$
So that
$$
mathbb{V}{rm ar}[X] = int_{-infty}^{+infty} {rm d}x~ x^2 f(x) - left(
int_{-infty}^{+infty} {rm d}x~ x f(x) right)^2 = a(sigma_1 - sigma_2)(sigma_1 + sigma_2) + sigma_2^2 + a(1 - a)(mu_1 - mu_2)^2 tag{3}
$$
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
$f_X$ is the density of the random variable $UZ_1+(1-U)Z_2$ where $Usim Bernoulli(0.8), Z_1sim N(50,25)$ and $Z_2sim N(60,64)$ and they are mutually independent.
Then $Var(UZ_1+(1-U)Z_2)=E[Var(UZ_1+(1-U)Z_2|U)]+Var[E(UZ_1+(1-U)Z_2|U)].$
$E[Var(UZ_1+(1-U)Z_2|U)]=E[U^2Var(Z_1|U)+(1-U)^2Var(Z_2|U)]=E[U^2]Var(Z_1)+E[(1-U)^2]Var(Z_2)=E[U]times 25+E[1-U]times 64=0.8times 25+0.2times 64=32.8.$
$Var[E(UZ_1+(1-U)Z_2|U)]=Var[UE(Z_1)+(1-U)E(Z_2)]=Var[50U+60(1-U)]=Var[-10U]=100times 0.8times 0.2=16.$
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
$endgroup$
add a comment |
$begingroup$
$f_X$ is the density of the random variable $UZ_1+(1-U)Z_2$ where $Usim Bernoulli(0.8), Z_1sim N(50,25)$ and $Z_2sim N(60,64)$ and they are mutually independent.
Then $Var(UZ_1+(1-U)Z_2)=E[Var(UZ_1+(1-U)Z_2|U)]+Var[E(UZ_1+(1-U)Z_2|U)].$
$E[Var(UZ_1+(1-U)Z_2|U)]=E[U^2Var(Z_1|U)+(1-U)^2Var(Z_2|U)]=E[U^2]Var(Z_1)+E[(1-U)^2]Var(Z_2)=E[U]times 25+E[1-U]times 64=0.8times 25+0.2times 64=32.8.$
$Var[E(UZ_1+(1-U)Z_2|U)]=Var[UE(Z_1)+(1-U)E(Z_2)]=Var[50U+60(1-U)]=Var[-10U]=100times 0.8times 0.2=16.$
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
$endgroup$
add a comment |
$begingroup$
$f_X$ is the density of the random variable $UZ_1+(1-U)Z_2$ where $Usim Bernoulli(0.8), Z_1sim N(50,25)$ and $Z_2sim N(60,64)$ and they are mutually independent.
Then $Var(UZ_1+(1-U)Z_2)=E[Var(UZ_1+(1-U)Z_2|U)]+Var[E(UZ_1+(1-U)Z_2|U)].$
$E[Var(UZ_1+(1-U)Z_2|U)]=E[U^2Var(Z_1|U)+(1-U)^2Var(Z_2|U)]=E[U^2]Var(Z_1)+E[(1-U)^2]Var(Z_2)=E[U]times 25+E[1-U]times 64=0.8times 25+0.2times 64=32.8.$
$Var[E(UZ_1+(1-U)Z_2|U)]=Var[UE(Z_1)+(1-U)E(Z_2)]=Var[50U+60(1-U)]=Var[-10U]=100times 0.8times 0.2=16.$
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
$endgroup$
$f_X$ is the density of the random variable $UZ_1+(1-U)Z_2$ where $Usim Bernoulli(0.8), Z_1sim N(50,25)$ and $Z_2sim N(60,64)$ and they are mutually independent.
Then $Var(UZ_1+(1-U)Z_2)=E[Var(UZ_1+(1-U)Z_2|U)]+Var[E(UZ_1+(1-U)Z_2|U)].$
$E[Var(UZ_1+(1-U)Z_2|U)]=E[U^2Var(Z_1|U)+(1-U)^2Var(Z_2|U)]=E[U^2]Var(Z_1)+E[(1-U)^2]Var(Z_2)=E[U]times 25+E[1-U]times 64=0.8times 25+0.2times 64=32.8.$
$Var[E(UZ_1+(1-U)Z_2|U)]=Var[UE(Z_1)+(1-U)E(Z_2)]=Var[50U+60(1-U)]=Var[-10U]=100times 0.8times 0.2=16.$
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
answered Dec 16 '18 at 23:30
John_WickJohn_Wick
1,616111
1,616111
add a comment |
add a comment |
$begingroup$
for a function of the form
$$
f(x) = a frac{1}{sqrt{2pi sigma_1^2}}e^{(x - mu_1)^2/2sigma_1^2} + (1- a)frac{1}{sqrt{2pi sigma_2^2}}e^{(x - mu_2)^2/2sigma_2^2}
$$
you have
$$
int_{-infty}^{+infty} {rm d}x~ f(x) = a + (1 - a) = 1
$$
$$
int_{-infty}^{+infty} {rm d}x~ x f(x) = amu_1 + (1 - a)mu_2 tag{1}
$$
and
$$
int_{-infty}^{+infty} {rm d}x~ x^2 f(x) = a(mu_1^2 + sigma_1^2) + (1 - a)(mu_2^2 + sigma_2^2) tag{2}
$$
So that
$$
mathbb{V}{rm ar}[X] = int_{-infty}^{+infty} {rm d}x~ x^2 f(x) - left(
int_{-infty}^{+infty} {rm d}x~ x f(x) right)^2 = a(sigma_1 - sigma_2)(sigma_1 + sigma_2) + sigma_2^2 + a(1 - a)(mu_1 - mu_2)^2 tag{3}
$$
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
for a function of the form
$$
f(x) = a frac{1}{sqrt{2pi sigma_1^2}}e^{(x - mu_1)^2/2sigma_1^2} + (1- a)frac{1}{sqrt{2pi sigma_2^2}}e^{(x - mu_2)^2/2sigma_2^2}
$$
you have
$$
int_{-infty}^{+infty} {rm d}x~ f(x) = a + (1 - a) = 1
$$
$$
int_{-infty}^{+infty} {rm d}x~ x f(x) = amu_1 + (1 - a)mu_2 tag{1}
$$
and
$$
int_{-infty}^{+infty} {rm d}x~ x^2 f(x) = a(mu_1^2 + sigma_1^2) + (1 - a)(mu_2^2 + sigma_2^2) tag{2}
$$
So that
$$
mathbb{V}{rm ar}[X] = int_{-infty}^{+infty} {rm d}x~ x^2 f(x) - left(
int_{-infty}^{+infty} {rm d}x~ x f(x) right)^2 = a(sigma_1 - sigma_2)(sigma_1 + sigma_2) + sigma_2^2 + a(1 - a)(mu_1 - mu_2)^2 tag{3}
$$
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
$endgroup$
add a comment |
$begingroup$
for a function of the form
$$
f(x) = a frac{1}{sqrt{2pi sigma_1^2}}e^{(x - mu_1)^2/2sigma_1^2} + (1- a)frac{1}{sqrt{2pi sigma_2^2}}e^{(x - mu_2)^2/2sigma_2^2}
$$
you have
$$
int_{-infty}^{+infty} {rm d}x~ f(x) = a + (1 - a) = 1
$$
$$
int_{-infty}^{+infty} {rm d}x~ x f(x) = amu_1 + (1 - a)mu_2 tag{1}
$$
and
$$
int_{-infty}^{+infty} {rm d}x~ x^2 f(x) = a(mu_1^2 + sigma_1^2) + (1 - a)(mu_2^2 + sigma_2^2) tag{2}
$$
So that
$$
mathbb{V}{rm ar}[X] = int_{-infty}^{+infty} {rm d}x~ x^2 f(x) - left(
int_{-infty}^{+infty} {rm d}x~ x f(x) right)^2 = a(sigma_1 - sigma_2)(sigma_1 + sigma_2) + sigma_2^2 + a(1 - a)(mu_1 - mu_2)^2 tag{3}
$$
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
$endgroup$
for a function of the form
$$
f(x) = a frac{1}{sqrt{2pi sigma_1^2}}e^{(x - mu_1)^2/2sigma_1^2} + (1- a)frac{1}{sqrt{2pi sigma_2^2}}e^{(x - mu_2)^2/2sigma_2^2}
$$
you have
$$
int_{-infty}^{+infty} {rm d}x~ f(x) = a + (1 - a) = 1
$$
$$
int_{-infty}^{+infty} {rm d}x~ x f(x) = amu_1 + (1 - a)mu_2 tag{1}
$$
and
$$
int_{-infty}^{+infty} {rm d}x~ x^2 f(x) = a(mu_1^2 + sigma_1^2) + (1 - a)(mu_2^2 + sigma_2^2) tag{2}
$$
So that
$$
mathbb{V}{rm ar}[X] = int_{-infty}^{+infty} {rm d}x~ x^2 f(x) - left(
int_{-infty}^{+infty} {rm d}x~ x f(x) right)^2 = a(sigma_1 - sigma_2)(sigma_1 + sigma_2) + sigma_2^2 + a(1 - a)(mu_1 - mu_2)^2 tag{3}
$$
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
answered Dec 16 '18 at 22:52
caveraccaverac
14.8k31130
14.8k31130
add a comment |
add a comment |
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