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$X$ ~ $N(u, sigma^2)$. Find $E(X)$
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$X$ ~ $N(u, sigma^2)$. Find $E(X)$
We know that $f_X(x) = frac1sigmasqrt2pie^-frac12left(fracx-usigma right)^2$
Also,
$Z = fracX-usigma$ ~ $N(0, 1)$, so $f_Z(z) = frac1sqrt2pie^-frac12z^2$
$E(Z) = Eleft[fracXsigma - fracusigma right] = frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$
Therefore $E(X) = u$ for $X$ ~ $N(u, sigma^2)$
I don't get it. The part where "$frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$"
I'm aware that $X$ ~ $N(0, 1)$, $E(X) = 0$.
probability expectation
asked 3 hours ago
Bas bas
1878
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up vote
0
down vote
favorite
$X$ ~ $N(u, sigma^2)$. Find $E(X)$
We know that $f_X(x) = frac1sigmasqrt2pie^-frac12left(fracx-usigma right)^2$
Also,
$Z = fracX-usigma$ ~ $N(0, 1)$, so $f_Z(z) = frac1sqrt2pie^-frac12z^2$
$E(Z) = Eleft[fracXsigma - fracusigma right] = frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$
Therefore $E(X) = u$ for $X$ ~ $N(u, sigma^2)$
I don't get it. The part where "$frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$"
I'm aware that $X$ ~ $N(0, 1)$, $E(X) = 0$.
probability expectation
asked 3 hours ago
Bas bas
1878
The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$X$ ~ $N(u, sigma^2)$. Find $E(X)$
We know that $f_X(x) = frac1sigmasqrt2pie^-frac12left(fracx-usigma right)^2$
Also,
$Z = fracX-usigma$ ~ $N(0, 1)$, so $f_Z(z) = frac1sqrt2pie^-frac12z^2$
$E(Z) = Eleft[fracXsigma - fracusigma right] = frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$
Therefore $E(X) = u$ for $X$ ~ $N(u, sigma^2)$
I don't get it. The part where "$frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$"
I'm aware that $X$ ~ $N(0, 1)$, $E(X) = 0$.
probability expectation
asked 3 hours ago
Bas bas
1878
$X$ ~ $N(u, sigma^2)$. Find $E(X)$
We know that $f_X(x) = frac1sigmasqrt2pie^-frac12left(fracx-usigma right)^2$
Also,
$Z = fracX-usigma$ ~ $N(0, 1)$, so $f_Z(z) = frac1sqrt2pie^-frac12z^2$
$E(Z) = Eleft[fracXsigma - fracusigma right] = frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$
Therefore $E(X) = u$ for $X$ ~ $N(u, sigma^2)$
I don't get it. The part where "$frac1sigmaE(X) + fracusigmaE(1) = E(X) = u$"
I'm aware that $X$ ~ $N(0, 1)$, $E(X) = 0$.
probability expectation
asked 3 hours ago
Bas bas
1878
edited 3 hours ago
asked 3 hours ago
Bas bas
1878
asked 3 hours ago
Bas bas
1878
asked 3 hours ago
Bas bas
1878
1878
The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago
add a comment |Â
The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago
The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago
add a comment |Â
1 Answer
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You have a very big mistake. First of all, the expectation operator is linear, therefore
$$E(Z) = frac1sigmaE(X) - fracmusigmaE(1) = fracmusigma-fracmusigma =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title:
$$E(X) = intlimits_-infty^+infty x f_X(x)$$
where
$$f_X(x) = frac1sigmasqrt2pie^-frac12sigma^2(x-mu)^2$$
So
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty x e^-frac12sigma^2(x-mu)^2 dx$$
Take a change of variable $t = x-mu$ so $dt = dx$ and then
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty (t+mu) e^-frac12sigma^2t^2 dt = frac1sigmasqrt2pi intlimits_-infty^+infty t e^-frac12sigma^2t^2 dt+mu intlimits_-infty^+infty frac1sigmasqrt2pie^-frac12sigma^2t^2 dt$$
So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally
$$E(X) = 0 + mu(1) = mu$$
answered 3 hours ago
Ahmad Bazzi
2,162416
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have a very big mistake. First of all, the expectation operator is linear, therefore
$$E(Z) = frac1sigmaE(X) - fracmusigmaE(1) = fracmusigma-fracmusigma =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title:
$$E(X) = intlimits_-infty^+infty x f_X(x)$$
where
$$f_X(x) = frac1sigmasqrt2pie^-frac12sigma^2(x-mu)^2$$
So
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty x e^-frac12sigma^2(x-mu)^2 dx$$
Take a change of variable $t = x-mu$ so $dt = dx$ and then
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty (t+mu) e^-frac12sigma^2t^2 dt = frac1sigmasqrt2pi intlimits_-infty^+infty t e^-frac12sigma^2t^2 dt+mu intlimits_-infty^+infty frac1sigmasqrt2pie^-frac12sigma^2t^2 dt$$
So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally
$$E(X) = 0 + mu(1) = mu$$
answered 3 hours ago
Ahmad Bazzi
2,162416
add a comment |Â
up vote
1
down vote
accepted
You have a very big mistake. First of all, the expectation operator is linear, therefore
$$E(Z) = frac1sigmaE(X) - fracmusigmaE(1) = fracmusigma-fracmusigma =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title:
$$E(X) = intlimits_-infty^+infty x f_X(x)$$
where
$$f_X(x) = frac1sigmasqrt2pie^-frac12sigma^2(x-mu)^2$$
So
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty x e^-frac12sigma^2(x-mu)^2 dx$$
Take a change of variable $t = x-mu$ so $dt = dx$ and then
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty (t+mu) e^-frac12sigma^2t^2 dt = frac1sigmasqrt2pi intlimits_-infty^+infty t e^-frac12sigma^2t^2 dt+mu intlimits_-infty^+infty frac1sigmasqrt2pie^-frac12sigma^2t^2 dt$$
So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally
$$E(X) = 0 + mu(1) = mu$$
answered 3 hours ago
Ahmad Bazzi
2,162416
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have a very big mistake. First of all, the expectation operator is linear, therefore
$$E(Z) = frac1sigmaE(X) - fracmusigmaE(1) = fracmusigma-fracmusigma =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title:
$$E(X) = intlimits_-infty^+infty x f_X(x)$$
where
$$f_X(x) = frac1sigmasqrt2pie^-frac12sigma^2(x-mu)^2$$
So
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty x e^-frac12sigma^2(x-mu)^2 dx$$
Take a change of variable $t = x-mu$ so $dt = dx$ and then
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty (t+mu) e^-frac12sigma^2t^2 dt = frac1sigmasqrt2pi intlimits_-infty^+infty t e^-frac12sigma^2t^2 dt+mu intlimits_-infty^+infty frac1sigmasqrt2pie^-frac12sigma^2t^2 dt$$
So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally
$$E(X) = 0 + mu(1) = mu$$
answered 3 hours ago
Ahmad Bazzi
2,162416
You have a very big mistake. First of all, the expectation operator is linear, therefore
$$E(Z) = frac1sigmaE(X) - fracmusigmaE(1) = fracmusigma-fracmusigma =0 $$
Second of all, the title's content is different from what you are asking. So from hereon, I will address the question in the title:
$$E(X) = intlimits_-infty^+infty x f_X(x)$$
where
$$f_X(x) = frac1sigmasqrt2pie^-frac12sigma^2(x-mu)^2$$
So
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty x e^-frac12sigma^2(x-mu)^2 dx$$
Take a change of variable $t = x-mu$ so $dt = dx$ and then
$$E(X) = frac1sigmasqrt2pi intlimits_-infty^+infty (t+mu) e^-frac12sigma^2t^2 dt = frac1sigmasqrt2pi intlimits_-infty^+infty t e^-frac12sigma^2t^2 dt+mu intlimits_-infty^+infty frac1sigmasqrt2pie^-frac12sigma^2t^2 dt$$
So the first integral evaluates to zero because it is an odd function. The second integral is the integration of the Normal PDF hence it evaluates to 1. Finally
$$E(X) = 0 + mu(1) = mu$$
answered 3 hours ago
Ahmad Bazzi
2,162416
answered 3 hours ago
Ahmad Bazzi
2,162416
answered 3 hours ago
Ahmad Bazzi
2,162416
answered 3 hours ago
Ahmad Bazzi
2,162416
2,162416
add a comment |Â
add a comment |Â
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The Line $E(Z)=E[fracXsigma-fracusigma]$ is wrong. The rest of the line should be $frac1sigmaE(X)-fracusigma=0$.
– herb steinberg
3 hours ago
Why not perform the integration $int_-infty^inftyxf_X(x),dx$?
– StubbornAtom
3 hours ago