independent, identically distributed (IID) random variables
I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.
Would someone explain in simple terms what IID random variables are and give me an example?
probability probability-theory random-variables independence
add a comment |
I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.
Would someone explain in simple terms what IID random variables are and give me an example?
probability probability-theory random-variables independence
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41
add a comment |
I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.
Would someone explain in simple terms what IID random variables are and give me an example?
probability probability-theory random-variables independence
I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.
Would someone explain in simple terms what IID random variables are and give me an example?
probability probability-theory random-variables independence
probability probability-theory random-variables independence
edited Nov 12 '15 at 9:25
BCLC
1
1
asked Aug 13 '13 at 21:35
Frank EppsFrank Epps
3321323
3321323
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41
add a comment |
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41
add a comment |
3 Answers
3
active
oldest
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Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.
If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.
They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.
They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).
Does that help?
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
add a comment |
"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$
For example, toss 2 dice. Let $X_1$ be the indicator RV of the first being {1, 2}, and let $X_2$ be the indicator RV of the second being {6}. It's intuitive to conclude that $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$, so are other combinations.
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be the indicator RV of the second die being {1, 6}. Then we get $P(X_1=1) = P(X_2=1)=frac{1}{3}$.
"Independent but not identical" as shown in the above two instances, the first is this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$ but $P(X_1) neq P(X_2)$.
"Identical and independent", the second example is for this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1)$ and $P(X_1) = P(X_2)$.
"Identical but not independent", let $X_2$ be the indicator RV of the first die being {5, 6}, we can get: $P(X_1=1, X_2 = 1) = 0 neq P(X_1=1) P(X_2=1)$ but $P(X_1) = P(X_2)$.
Refrence: https://math.stackexchange.com/a/994136/351322
add a comment |
.So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
add a comment |
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3 Answers
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3 Answers
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Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.
If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.
They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.
They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).
Does that help?
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
add a comment |
Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.
If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.
They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.
They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).
Does that help?
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
add a comment |
Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.
If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.
They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.
They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).
Does that help?
Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.
If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.
They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.
They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).
Does that help?
edited Aug 13 '13 at 23:41
answered Aug 13 '13 at 21:41
Dima McGreenDima McGreen
550412
550412
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
add a comment |
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
An example of an identical dependent situation would help me :).
– jiggunjer
Sep 5 '16 at 5:46
1
1
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
@jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
– Steven Stadnicki
Nov 6 '16 at 2:41
add a comment |
"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$
For example, toss 2 dice. Let $X_1$ be the indicator RV of the first being {1, 2}, and let $X_2$ be the indicator RV of the second being {6}. It's intuitive to conclude that $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$, so are other combinations.
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be the indicator RV of the second die being {1, 6}. Then we get $P(X_1=1) = P(X_2=1)=frac{1}{3}$.
"Independent but not identical" as shown in the above two instances, the first is this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$ but $P(X_1) neq P(X_2)$.
"Identical and independent", the second example is for this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1)$ and $P(X_1) = P(X_2)$.
"Identical but not independent", let $X_2$ be the indicator RV of the first die being {5, 6}, we can get: $P(X_1=1, X_2 = 1) = 0 neq P(X_1=1) P(X_2=1)$ but $P(X_1) = P(X_2)$.
Refrence: https://math.stackexchange.com/a/994136/351322
add a comment |
"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$
For example, toss 2 dice. Let $X_1$ be the indicator RV of the first being {1, 2}, and let $X_2$ be the indicator RV of the second being {6}. It's intuitive to conclude that $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$, so are other combinations.
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be the indicator RV of the second die being {1, 6}. Then we get $P(X_1=1) = P(X_2=1)=frac{1}{3}$.
"Independent but not identical" as shown in the above two instances, the first is this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$ but $P(X_1) neq P(X_2)$.
"Identical and independent", the second example is for this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1)$ and $P(X_1) = P(X_2)$.
"Identical but not independent", let $X_2$ be the indicator RV of the first die being {5, 6}, we can get: $P(X_1=1, X_2 = 1) = 0 neq P(X_1=1) P(X_2=1)$ but $P(X_1) = P(X_2)$.
Refrence: https://math.stackexchange.com/a/994136/351322
add a comment |
"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$
For example, toss 2 dice. Let $X_1$ be the indicator RV of the first being {1, 2}, and let $X_2$ be the indicator RV of the second being {6}. It's intuitive to conclude that $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$, so are other combinations.
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be the indicator RV of the second die being {1, 6}. Then we get $P(X_1=1) = P(X_2=1)=frac{1}{3}$.
"Independent but not identical" as shown in the above two instances, the first is this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$ but $P(X_1) neq P(X_2)$.
"Identical and independent", the second example is for this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1)$ and $P(X_1) = P(X_2)$.
"Identical but not independent", let $X_2$ be the indicator RV of the first die being {5, 6}, we can get: $P(X_1=1, X_2 = 1) = 0 neq P(X_1=1) P(X_2=1)$ but $P(X_1) = P(X_2)$.
Refrence: https://math.stackexchange.com/a/994136/351322
"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$
For example, toss 2 dice. Let $X_1$ be the indicator RV of the first being {1, 2}, and let $X_2$ be the indicator RV of the second being {6}. It's intuitive to conclude that $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$, so are other combinations.
However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be the indicator RV of the second die being {1, 6}. Then we get $P(X_1=1) = P(X_2=1)=frac{1}{3}$.
"Independent but not identical" as shown in the above two instances, the first is this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1) = frac{1}{3} cdot frac{1}{6}$ but $P(X_1) neq P(X_2)$.
"Identical and independent", the second example is for this case, since $P(X_1=1, X_2 = 1) = P(X_1=1) P(X_2=1)$ and $P(X_1) = P(X_2)$.
"Identical but not independent", let $X_2$ be the indicator RV of the first die being {5, 6}, we can get: $P(X_1=1, X_2 = 1) = 0 neq P(X_1=1) P(X_2=1)$ but $P(X_1) = P(X_2)$.
Refrence: https://math.stackexchange.com/a/994136/351322
edited Dec 5 '18 at 12:13
Community♦
1
1
answered Nov 6 '16 at 2:24
lernerlerner
289115
289115
add a comment |
add a comment |
.So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
add a comment |
.So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
add a comment |
.So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer
.So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer
answered Mar 11 '16 at 14:26
Tejas SureshTejas Suresh
111
111
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
add a comment |
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
– Shailesh
Mar 11 '16 at 14:46
add a comment |
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say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
– Eleven-Eleven
Aug 13 '13 at 21:41