How can I find the expectation and variance of $Z=max{X,Y}$ where $X$ and $Y$ are defined through joint...
$begingroup$
Random variables $X$ and $Y$ and have the joint distribution below, and $Z=max{X,Y}$
$$
begin{array}{c|lcr}
text{XY} & text{1} & text{2} & text{3} \
hline
1 & 0.12 & 0.08 & 0.20 \
2 & 0.18 & 0.12 & 0.30 \
end{array}
$$
Find $E[Z]$ and $V[Z]$
I am unable to understand that if $Z=max{X,Y}$ then how will we take the pairs? Or just $Y=3$ ? Because it is the only maximum. Please explain I am just stuck here.
probability statistics random-variables variance expected-value
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
$endgroup$
add a comment |
$begingroup$
Random variables $X$ and $Y$ and have the joint distribution below, and $Z=max{X,Y}$
$$
begin{array}{c|lcr}
text{XY} & text{1} & text{2} & text{3} \
hline
1 & 0.12 & 0.08 & 0.20 \
2 & 0.18 & 0.12 & 0.30 \
end{array}
$$
Find $E[Z]$ and $V[Z]$
I am unable to understand that if $Z=max{X,Y}$ then how will we take the pairs? Or just $Y=3$ ? Because it is the only maximum. Please explain I am just stuck here.
probability statistics random-variables variance expected-value
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
$endgroup$
add a comment |
$begingroup$
Random variables $X$ and $Y$ and have the joint distribution below, and $Z=max{X,Y}$
$$
begin{array}{c|lcr}
text{XY} & text{1} & text{2} & text{3} \
hline
1 & 0.12 & 0.08 & 0.20 \
2 & 0.18 & 0.12 & 0.30 \
end{array}
$$
Find $E[Z]$ and $V[Z]$
I am unable to understand that if $Z=max{X,Y}$ then how will we take the pairs? Or just $Y=3$ ? Because it is the only maximum. Please explain I am just stuck here.
probability statistics random-variables variance expected-value
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
$endgroup$
Random variables $X$ and $Y$ and have the joint distribution below, and $Z=max{X,Y}$
$$
begin{array}{c|lcr}
text{XY} & text{1} & text{2} & text{3} \
hline
1 & 0.12 & 0.08 & 0.20 \
2 & 0.18 & 0.12 & 0.30 \
end{array}
$$
Find $E[Z]$ and $V[Z]$
I am unable to understand that if $Z=max{X,Y}$ then how will we take the pairs? Or just $Y=3$ ? Because it is the only maximum. Please explain I am just stuck here.
probability statistics random-variables variance expected-value
probability statistics random-variables variance expected-value
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
edited Dec 13 '18 at 19:07
Yanior Weg
1,1451935
1,1451935
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
asked Dec 13 '18 at 16:26
Kriti AroraKriti Arora
396
396
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $Z = max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $Z = max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
$endgroup$
add a comment |
$begingroup$
If $Z = max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
$endgroup$
add a comment |
$begingroup$
If $Z = max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
$endgroup$
If $Z = max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.12 + 0.76 + 1.5 = 2.38$
$E(Z^2) = 0.12 + 4*0.38 + 9*0.5 = 0.12 + 1.52+ 4.5 = 6.04$
That means $V(Z) = E(Z^2) - (E(Z))^2 = 6.04 - 5.6644 = 0.3756$.
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
answered Dec 13 '18 at 18:26
Yanior WegYanior Weg
1,1451935
1,1451935
add a comment |
add a comment |
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