CDF of the of Joint Uniform Distribution for k random variables.
If i have K independent Random Variable:
$X_1,X_2,x_3,cdotcdotcdotcdotcdotcdotcdotcdotcdot, X_k$
What would be the CDF of the sum of their Joint Distribution?
$f_{X_1+X_2+X_3+...+X_k} (z)$ z<=1.
I cant figure out what would be their respective density functions and integration limits.
probability probability-distributions uniform-distribution
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
add a comment |
If i have K independent Random Variable:
$X_1,X_2,x_3,cdotcdotcdotcdotcdotcdotcdotcdotcdot, X_k$
What would be the CDF of the sum of their Joint Distribution?
$f_{X_1+X_2+X_3+...+X_k} (z)$ z<=1.
I cant figure out what would be their respective density functions and integration limits.
probability probability-distributions uniform-distribution
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
add a comment |
If i have K independent Random Variable:
$X_1,X_2,x_3,cdotcdotcdotcdotcdotcdotcdotcdotcdot, X_k$
What would be the CDF of the sum of their Joint Distribution?
$f_{X_1+X_2+X_3+...+X_k} (z)$ z<=1.
I cant figure out what would be their respective density functions and integration limits.
probability probability-distributions uniform-distribution
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
If i have K independent Random Variable:
$X_1,X_2,x_3,cdotcdotcdotcdotcdotcdotcdotcdotcdot, X_k$
What would be the CDF of the sum of their Joint Distribution?
$f_{X_1+X_2+X_3+...+X_k} (z)$ z<=1.
I cant figure out what would be their respective density functions and integration limits.
probability probability-distributions uniform-distribution
probability probability-distributions uniform-distribution
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
edited Nov 18 at 14:37
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
asked Nov 18 at 14:04
Muhammad Fasiurrehman Sohi
187
187
add a comment |
add a comment |
2 Answers
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By independence:$$F_{X_1,dots,X_k}(x_1,dots,x_k)=P(X_1leq x_1,dots,X_kleq x_k)=$$$$P(X_1leq x_1)timescdotstimes P(X_kleq x_k)=F_{X_1}(x_1)timescdotstimes F_{X_k}(x_k)$$
answered Nov 18 at 14:10
drhab
97.8k544129
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If they are independent, then
$$Pr(X_1 le x_1, ldots, X_k le x_k ) = prod_{i=1}^k Pr(X_i le x_i)$$
That is the joint CDF is just the product of individual CDF.
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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By independence:$$F_{X_1,dots,X_k}(x_1,dots,x_k)=P(X_1leq x_1,dots,X_kleq x_k)=$$$$P(X_1leq x_1)timescdotstimes P(X_kleq x_k)=F_{X_1}(x_1)timescdotstimes F_{X_k}(x_k)$$
answered Nov 18 at 14:10
drhab
97.8k544129
add a comment |
By independence:$$F_{X_1,dots,X_k}(x_1,dots,x_k)=P(X_1leq x_1,dots,X_kleq x_k)=$$$$P(X_1leq x_1)timescdotstimes P(X_kleq x_k)=F_{X_1}(x_1)timescdotstimes F_{X_k}(x_k)$$
answered Nov 18 at 14:10
drhab
97.8k544129
add a comment |
By independence:$$F_{X_1,dots,X_k}(x_1,dots,x_k)=P(X_1leq x_1,dots,X_kleq x_k)=$$$$P(X_1leq x_1)timescdotstimes P(X_kleq x_k)=F_{X_1}(x_1)timescdotstimes F_{X_k}(x_k)$$
answered Nov 18 at 14:10
drhab
97.8k544129
By independence:$$F_{X_1,dots,X_k}(x_1,dots,x_k)=P(X_1leq x_1,dots,X_kleq x_k)=$$$$P(X_1leq x_1)timescdotstimes P(X_kleq x_k)=F_{X_1}(x_1)timescdotstimes F_{X_k}(x_k)$$
answered Nov 18 at 14:10
drhab
97.8k544129
answered Nov 18 at 14:10
drhab
97.8k544129
answered Nov 18 at 14:10
drhab
97.8k544129
answered Nov 18 at 14:10
drhab
97.8k544129
97.8k544129
add a comment |
add a comment |
If they are independent, then
$$Pr(X_1 le x_1, ldots, X_k le x_k ) = prod_{i=1}^k Pr(X_i le x_i)$$
That is the joint CDF is just the product of individual CDF.
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
add a comment |
If they are independent, then
$$Pr(X_1 le x_1, ldots, X_k le x_k ) = prod_{i=1}^k Pr(X_i le x_i)$$
That is the joint CDF is just the product of individual CDF.
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
add a comment |
If they are independent, then
$$Pr(X_1 le x_1, ldots, X_k le x_k ) = prod_{i=1}^k Pr(X_i le x_i)$$
That is the joint CDF is just the product of individual CDF.
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
If they are independent, then
$$Pr(X_1 le x_1, ldots, X_k le x_k ) = prod_{i=1}^k Pr(X_i le x_i)$$
That is the joint CDF is just the product of individual CDF.
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
answered Nov 18 at 14:10
Siong Thye Goh
99.3k1464117
99.3k1464117
add a comment |
add a comment |
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