Expectation and variance of travel time with several options for the transportation
$begingroup$
A person is traveling between two places, and has 3 options for transportation. The jth option would take an average of µj hours, with a standard deviation of $sigma_j$ hours. The person randomly chooses between the 3 options, with equal probabilities. Let T be how long it takes for
him to get from place 1 to place 2.
(a) Find E(T). Is it simply (µ1 +µ2 +µ3)/3, the average of the expectations?
Expectation is additive, there is equal probabilities between options, so
I agree it is the "average" of the averages.
(b) Find Var(T). Is it simply ($sigma_1^2+sigma_2^2+sigma_j^2$)/3, the average of the variances?
Variance is additive when you are adding iid E(T)'s (confirmed: https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/why-variances-add-and-why-it-matters), but I am unsure if you just "average" the variances. Guidance?
probability-theory random-variables variance expected-value
edited Dec 9 '18 at 9:49
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asked Dec 9 '18 at 2:03
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$begingroup$
A person is traveling between two places, and has 3 options for transportation. The jth option would take an average of µj hours, with a standard deviation of $sigma_j$ hours. The person randomly chooses between the 3 options, with equal probabilities. Let T be how long it takes for
him to get from place 1 to place 2.
(a) Find E(T). Is it simply (µ1 +µ2 +µ3)/3, the average of the expectations?
Expectation is additive, there is equal probabilities between options, so
I agree it is the "average" of the averages.
(b) Find Var(T). Is it simply ($sigma_1^2+sigma_2^2+sigma_j^2$)/3, the average of the variances?
Variance is additive when you are adding iid E(T)'s (confirmed: https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/why-variances-add-and-why-it-matters), but I am unsure if you just "average" the variances. Guidance?
probability-theory random-variables variance expected-value
edited Dec 9 '18 at 9:49
Did
247k23222458
asked Dec 9 '18 at 2:03
user603569user603569
728
$endgroup$
add a comment |
$begingroup$
A person is traveling between two places, and has 3 options for transportation. The jth option would take an average of µj hours, with a standard deviation of $sigma_j$ hours. The person randomly chooses between the 3 options, with equal probabilities. Let T be how long it takes for
him to get from place 1 to place 2.
(a) Find E(T). Is it simply (µ1 +µ2 +µ3)/3, the average of the expectations?
Expectation is additive, there is equal probabilities between options, so
I agree it is the "average" of the averages.
(b) Find Var(T). Is it simply ($sigma_1^2+sigma_2^2+sigma_j^2$)/3, the average of the variances?
Variance is additive when you are adding iid E(T)'s (confirmed: https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/why-variances-add-and-why-it-matters), but I am unsure if you just "average" the variances. Guidance?
probability-theory random-variables variance expected-value
edited Dec 9 '18 at 9:49
Did
247k23222458
asked Dec 9 '18 at 2:03
user603569user603569
728
$endgroup$
A person is traveling between two places, and has 3 options for transportation. The jth option would take an average of µj hours, with a standard deviation of $sigma_j$ hours. The person randomly chooses between the 3 options, with equal probabilities. Let T be how long it takes for
him to get from place 1 to place 2.
(a) Find E(T). Is it simply (µ1 +µ2 +µ3)/3, the average of the expectations?
Expectation is additive, there is equal probabilities between options, so
I agree it is the "average" of the averages.
(b) Find Var(T). Is it simply ($sigma_1^2+sigma_2^2+sigma_j^2$)/3, the average of the variances?
Variance is additive when you are adding iid E(T)'s (confirmed: https://apcentral.collegeboard.org/courses/ap-statistics/classroom-resources/why-variances-add-and-why-it-matters), but I am unsure if you just "average" the variances. Guidance?
probability-theory random-variables variance expected-value
probability-theory random-variables variance expected-value
edited Dec 9 '18 at 9:49
Did
247k23222458
asked Dec 9 '18 at 2:03
user603569user603569
728
edited Dec 9 '18 at 9:49
Did
247k23222458
asked Dec 9 '18 at 2:03
user603569user603569
728
edited Dec 9 '18 at 9:49
Did
247k23222458
edited Dec 9 '18 at 9:49
Did
247k23222458
edited Dec 9 '18 at 9:49
Did
247k23222458
247k23222458
asked Dec 9 '18 at 2:03
user603569user603569
728
asked Dec 9 '18 at 2:03
user603569user603569
728
asked Dec 9 '18 at 2:03
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1 Answer
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$begingroup$
Call $C$ the option chosen, thus $C$ is uniform on ${1,2,3}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
$$T=sum_jmathbf 1_{C=j}X_j$$
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
$$text{$C$ is independent of $(X_1,X_2,X_3)$}$$
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of ${X_1,X_2,X_3}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence,
$$E(T)=sum_jP(C=j)E(X_j)=frac13sum_jmu_j$$ as you predicted (but for a slightly different reason). Likewise,
$$T^2=sum_jmathbf 1_{C=j}X_j^2$$ hence, again by independence,
$$E(T^2)=sum_jP(C=j)E(X_j^2)=frac13sum_j(sigma_j^2+mu_j^2)$$ which implies that
$$mathrm{var}(T)=frac13sum_jsigma_j^2+frac13sum_jmu_j^2-left(frac13sum_jmu_jright)^2$$
More generally, for any given number of choices, each with probability $p_j$, one would get
$$E(T)=sum_jp_jmu_j$$ and $$mathrm{var}(T)=sum_jp_jsigma_j^2+sum_jp_jmu_j^2-left(sum_jp_jmu_jright)^2$$
answered Dec 9 '18 at 9:43
DidDid
247k23222458
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1 Answer
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$begingroup$
Call $C$ the option chosen, thus $C$ is uniform on ${1,2,3}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
$$T=sum_jmathbf 1_{C=j}X_j$$
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
$$text{$C$ is independent of $(X_1,X_2,X_3)$}$$
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of ${X_1,X_2,X_3}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence,
$$E(T)=sum_jP(C=j)E(X_j)=frac13sum_jmu_j$$ as you predicted (but for a slightly different reason). Likewise,
$$T^2=sum_jmathbf 1_{C=j}X_j^2$$ hence, again by independence,
$$E(T^2)=sum_jP(C=j)E(X_j^2)=frac13sum_j(sigma_j^2+mu_j^2)$$ which implies that
$$mathrm{var}(T)=frac13sum_jsigma_j^2+frac13sum_jmu_j^2-left(frac13sum_jmu_jright)^2$$
More generally, for any given number of choices, each with probability $p_j$, one would get
$$E(T)=sum_jp_jmu_j$$ and $$mathrm{var}(T)=sum_jp_jsigma_j^2+sum_jp_jmu_j^2-left(sum_jp_jmu_jright)^2$$
answered Dec 9 '18 at 9:43
DidDid
247k23222458
$endgroup$
add a comment |
$begingroup$
Call $C$ the option chosen, thus $C$ is uniform on ${1,2,3}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
$$T=sum_jmathbf 1_{C=j}X_j$$
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
$$text{$C$ is independent of $(X_1,X_2,X_3)$}$$
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of ${X_1,X_2,X_3}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence,
$$E(T)=sum_jP(C=j)E(X_j)=frac13sum_jmu_j$$ as you predicted (but for a slightly different reason). Likewise,
$$T^2=sum_jmathbf 1_{C=j}X_j^2$$ hence, again by independence,
$$E(T^2)=sum_jP(C=j)E(X_j^2)=frac13sum_j(sigma_j^2+mu_j^2)$$ which implies that
$$mathrm{var}(T)=frac13sum_jsigma_j^2+frac13sum_jmu_j^2-left(frac13sum_jmu_jright)^2$$
More generally, for any given number of choices, each with probability $p_j$, one would get
$$E(T)=sum_jp_jmu_j$$ and $$mathrm{var}(T)=sum_jp_jsigma_j^2+sum_jp_jmu_j^2-left(sum_jp_jmu_jright)^2$$
answered Dec 9 '18 at 9:43
DidDid
247k23222458
$endgroup$
add a comment |
$begingroup$
Call $C$ the option chosen, thus $C$ is uniform on ${1,2,3}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
$$T=sum_jmathbf 1_{C=j}X_j$$
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
$$text{$C$ is independent of $(X_1,X_2,X_3)$}$$
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of ${X_1,X_2,X_3}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence,
$$E(T)=sum_jP(C=j)E(X_j)=frac13sum_jmu_j$$ as you predicted (but for a slightly different reason). Likewise,
$$T^2=sum_jmathbf 1_{C=j}X_j^2$$ hence, again by independence,
$$E(T^2)=sum_jP(C=j)E(X_j^2)=frac13sum_j(sigma_j^2+mu_j^2)$$ which implies that
$$mathrm{var}(T)=frac13sum_jsigma_j^2+frac13sum_jmu_j^2-left(frac13sum_jmu_jright)^2$$
More generally, for any given number of choices, each with probability $p_j$, one would get
$$E(T)=sum_jp_jmu_j$$ and $$mathrm{var}(T)=sum_jp_jsigma_j^2+sum_jp_jmu_j^2-left(sum_jp_jmu_jright)^2$$
answered Dec 9 '18 at 9:43
DidDid
247k23222458
$endgroup$
Call $C$ the option chosen, thus $C$ is uniform on ${1,2,3}$. The random variable $C$ allows to give a rigorous, pathwise, description of $T$ as
$$T=sum_jmathbf 1_{C=j}X_j$$
where $X_j$ is the transportation time for option $j$.
Now, add to the pot the crucial assumption that:
$$text{$C$ is independent of $(X_1,X_2,X_3)$}$$
That is, one assumes that the passenger chooses their option without any knowledge about the transportation times they shall have to endure (otherwise, one could imagine that $T$ is the minimum of ${X_1,X_2,X_3}$...).
With this representation of $T$ and this independence hypothesis, we can compute every characteristic of $T$. For example, by independence,
$$E(T)=sum_jP(C=j)E(X_j)=frac13sum_jmu_j$$ as you predicted (but for a slightly different reason). Likewise,
$$T^2=sum_jmathbf 1_{C=j}X_j^2$$ hence, again by independence,
$$E(T^2)=sum_jP(C=j)E(X_j^2)=frac13sum_j(sigma_j^2+mu_j^2)$$ which implies that
$$mathrm{var}(T)=frac13sum_jsigma_j^2+frac13sum_jmu_j^2-left(frac13sum_jmu_jright)^2$$
More generally, for any given number of choices, each with probability $p_j$, one would get
$$E(T)=sum_jp_jmu_j$$ and $$mathrm{var}(T)=sum_jp_jsigma_j^2+sum_jp_jmu_j^2-left(sum_jp_jmu_jright)^2$$
answered Dec 9 '18 at 9:43
DidDid
247k23222458
answered Dec 9 '18 at 9:43
DidDid
247k23222458
answered Dec 9 '18 at 9:43
DidDid
247k23222458
answered Dec 9 '18 at 9:43
DidDid
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247k23222458
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