Expected value and variance of a transformed random variable
$begingroup$
X is a binomial random variable with $n = 10000$ and $p = 60%$ (hence $E(X) = 10000 * 60 = 6000$). Now Y is a transformed random variable with $Y = 100000 - 7X$.
I have to calculate the expected value $E(Y)$ and variance $V(Y)$ of Y.
For $E(Y)$ I thought of this: $E(Y) = E(100000 - 7X) = E(-7X) + 100000 = 100000 - E(7X) = 100000 - 7E(X) = 100000 - 7 * 6000 = 58000$
But how do I calculate $V(Y)$?
statistics random-variables variance expected-value
asked Dec 3 '18 at 8:38
JoeyJoey
436
$endgroup$
add a comment |
$begingroup$
X is a binomial random variable with $n = 10000$ and $p = 60%$ (hence $E(X) = 10000 * 60 = 6000$). Now Y is a transformed random variable with $Y = 100000 - 7X$.
I have to calculate the expected value $E(Y)$ and variance $V(Y)$ of Y.
For $E(Y)$ I thought of this: $E(Y) = E(100000 - 7X) = E(-7X) + 100000 = 100000 - E(7X) = 100000 - 7E(X) = 100000 - 7 * 6000 = 58000$
But how do I calculate $V(Y)$?
statistics random-variables variance expected-value
asked Dec 3 '18 at 8:38
JoeyJoey
436
$endgroup$
add a comment |
$begingroup$
X is a binomial random variable with $n = 10000$ and $p = 60%$ (hence $E(X) = 10000 * 60 = 6000$). Now Y is a transformed random variable with $Y = 100000 - 7X$.
I have to calculate the expected value $E(Y)$ and variance $V(Y)$ of Y.
For $E(Y)$ I thought of this: $E(Y) = E(100000 - 7X) = E(-7X) + 100000 = 100000 - E(7X) = 100000 - 7E(X) = 100000 - 7 * 6000 = 58000$
But how do I calculate $V(Y)$?
statistics random-variables variance expected-value
asked Dec 3 '18 at 8:38
JoeyJoey
436
$endgroup$
X is a binomial random variable with $n = 10000$ and $p = 60%$ (hence $E(X) = 10000 * 60 = 6000$). Now Y is a transformed random variable with $Y = 100000 - 7X$.
I have to calculate the expected value $E(Y)$ and variance $V(Y)$ of Y.
For $E(Y)$ I thought of this: $E(Y) = E(100000 - 7X) = E(-7X) + 100000 = 100000 - E(7X) = 100000 - 7E(X) = 100000 - 7 * 6000 = 58000$
But how do I calculate $V(Y)$?
statistics random-variables variance expected-value
statistics random-variables variance expected-value
asked Dec 3 '18 at 8:38
JoeyJoey
436
asked Dec 3 '18 at 8:38
JoeyJoey
436
asked Dec 3 '18 at 8:38
JoeyJoey
436
asked Dec 3 '18 at 8:38
JoeyJoey
436
asked Dec 3 '18 at 8:38
JoeyJoey
436
436
add a comment |
add a comment |
2 Answers
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$begingroup$
$$V(Y)=V(100000-7X)=7^2cdot V(X)$$
I believe you can finish the exercise from here.
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
In general if $a,b$ are constants then: $$mathsf{Var}(aX+b)=a^2mathsf{Var}(X)$$
Verification: $$mathbb Eleft(aX+b-mathbb E(aX+b)right)^2=mathbb Eleft(aX+b-amathbb EX-b)right)^2=mathbb Ea^2(X-mathbb EX)^2=a^2mathbb E(X-mathbb EX)^2$$
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
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2 Answers
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2 Answers
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$begingroup$
$$V(Y)=V(100000-7X)=7^2cdot V(X)$$
I believe you can finish the exercise from here.
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
$$V(Y)=V(100000-7X)=7^2cdot V(X)$$
I believe you can finish the exercise from here.
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
$$V(Y)=V(100000-7X)=7^2cdot V(X)$$
I believe you can finish the exercise from here.
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
$$V(Y)=V(100000-7X)=7^2cdot V(X)$$
I believe you can finish the exercise from here.
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 3 '18 at 8:41
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
add a comment |
add a comment |
$begingroup$
In general if $a,b$ are constants then: $$mathsf{Var}(aX+b)=a^2mathsf{Var}(X)$$
Verification: $$mathbb Eleft(aX+b-mathbb E(aX+b)right)^2=mathbb Eleft(aX+b-amathbb EX-b)right)^2=mathbb Ea^2(X-mathbb EX)^2=a^2mathbb E(X-mathbb EX)^2$$
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
$endgroup$
add a comment |
$begingroup$
In general if $a,b$ are constants then: $$mathsf{Var}(aX+b)=a^2mathsf{Var}(X)$$
Verification: $$mathbb Eleft(aX+b-mathbb E(aX+b)right)^2=mathbb Eleft(aX+b-amathbb EX-b)right)^2=mathbb Ea^2(X-mathbb EX)^2=a^2mathbb E(X-mathbb EX)^2$$
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
$endgroup$
add a comment |
$begingroup$
In general if $a,b$ are constants then: $$mathsf{Var}(aX+b)=a^2mathsf{Var}(X)$$
Verification: $$mathbb Eleft(aX+b-mathbb E(aX+b)right)^2=mathbb Eleft(aX+b-amathbb EX-b)right)^2=mathbb Ea^2(X-mathbb EX)^2=a^2mathbb E(X-mathbb EX)^2$$
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
$endgroup$
In general if $a,b$ are constants then: $$mathsf{Var}(aX+b)=a^2mathsf{Var}(X)$$
Verification: $$mathbb Eleft(aX+b-mathbb E(aX+b)right)^2=mathbb Eleft(aX+b-amathbb EX-b)right)^2=mathbb Ea^2(X-mathbb EX)^2=a^2mathbb E(X-mathbb EX)^2$$
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
answered Dec 3 '18 at 9:00
drhabdrhab
100k544130
100k544130
add a comment |
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