Mixing
Find $P(Y(x)gt E[Y(x)])$, where $Y(x)= min{i:X_igt x}$ for $(X_i)$ i.i.d.
$begingroup$
Let $X_i$ be i.i.d random variables with common cdf F. For a given constant x, define $Y(x)= min{i:X_igt x}$. Find $P(Y(x)gt E[Y(x)])$. What is the limit as $xtoinfty$?
My answer:
$P(Y(x)=1)=P(X_1gt x)=1-F$
$P(Y(x)=2)=P(X_1lt x,X_2gt x)=F(1-F)$
$P(Y(x)=3)=P(X_1lt x,X_2lt x,X_3gt x)=F^2(1-F)$
....
$P(Y(x)=k)=F^{k-1}(1-F)$
Thus, $E[Y(x)]=sum_{k=1}^{infty}kF^{k-1}(1-F)={1over 1-F}$
$P(Y(x)gt E[Y(x)])=P(Y(x)gt {1over 1-F})=sum_{k={*{1over 1-F}*}+1}^{infty}kF^{k-1}(1-F)$
where $*{1over 1-F}*$ denote the largest integer less than ${1over 1-F}$
Since $F=1$ when $x=infty$, $E(Y(x))=infty$, $P(Y(x)gt E[Y(x)])=0$
Please correct me if I am wrong.
probability-theory expected-value
edited Jan 20 at 16:48
Did
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
$endgroup$
add a comment |
$begingroup$
Let $X_i$ be i.i.d random variables with common cdf F. For a given constant x, define $Y(x)= min{i:X_igt x}$. Find $P(Y(x)gt E[Y(x)])$. What is the limit as $xtoinfty$?
My answer:
$P(Y(x)=1)=P(X_1gt x)=1-F$
$P(Y(x)=2)=P(X_1lt x,X_2gt x)=F(1-F)$
$P(Y(x)=3)=P(X_1lt x,X_2lt x,X_3gt x)=F^2(1-F)$
....
$P(Y(x)=k)=F^{k-1}(1-F)$
Thus, $E[Y(x)]=sum_{k=1}^{infty}kF^{k-1}(1-F)={1over 1-F}$
$P(Y(x)gt E[Y(x)])=P(Y(x)gt {1over 1-F})=sum_{k={*{1over 1-F}*}+1}^{infty}kF^{k-1}(1-F)$
where $*{1over 1-F}*$ denote the largest integer less than ${1over 1-F}$
Since $F=1$ when $x=infty$, $E(Y(x))=infty$, $P(Y(x)gt E[Y(x)])=0$
Please correct me if I am wrong.
probability-theory expected-value
edited Jan 20 at 16:48
Did
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
$endgroup$
$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53
add a comment |
$begingroup$
Let $X_i$ be i.i.d random variables with common cdf F. For a given constant x, define $Y(x)= min{i:X_igt x}$. Find $P(Y(x)gt E[Y(x)])$. What is the limit as $xtoinfty$?
My answer:
$P(Y(x)=1)=P(X_1gt x)=1-F$
$P(Y(x)=2)=P(X_1lt x,X_2gt x)=F(1-F)$
$P(Y(x)=3)=P(X_1lt x,X_2lt x,X_3gt x)=F^2(1-F)$
....
$P(Y(x)=k)=F^{k-1}(1-F)$
Thus, $E[Y(x)]=sum_{k=1}^{infty}kF^{k-1}(1-F)={1over 1-F}$
$P(Y(x)gt E[Y(x)])=P(Y(x)gt {1over 1-F})=sum_{k={*{1over 1-F}*}+1}^{infty}kF^{k-1}(1-F)$
where $*{1over 1-F}*$ denote the largest integer less than ${1over 1-F}$
Since $F=1$ when $x=infty$, $E(Y(x))=infty$, $P(Y(x)gt E[Y(x)])=0$
Please correct me if I am wrong.
probability-theory expected-value
edited Jan 20 at 16:48
Did
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
$endgroup$
Let $X_i$ be i.i.d random variables with common cdf F. For a given constant x, define $Y(x)= min{i:X_igt x}$. Find $P(Y(x)gt E[Y(x)])$. What is the limit as $xtoinfty$?
My answer:
$P(Y(x)=1)=P(X_1gt x)=1-F$
$P(Y(x)=2)=P(X_1lt x,X_2gt x)=F(1-F)$
$P(Y(x)=3)=P(X_1lt x,X_2lt x,X_3gt x)=F^2(1-F)$
....
$P(Y(x)=k)=F^{k-1}(1-F)$
Thus, $E[Y(x)]=sum_{k=1}^{infty}kF^{k-1}(1-F)={1over 1-F}$
$P(Y(x)gt E[Y(x)])=P(Y(x)gt {1over 1-F})=sum_{k={*{1over 1-F}*}+1}^{infty}kF^{k-1}(1-F)$
where $*{1over 1-F}*$ denote the largest integer less than ${1over 1-F}$
Since $F=1$ when $x=infty$, $E(Y(x))=infty$, $P(Y(x)gt E[Y(x)])=0$
Please correct me if I am wrong.
probability-theory expected-value
probability-theory expected-value
edited Jan 20 at 16:48
Did
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
edited Jan 20 at 16:48
Did
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
edited Jan 20 at 16:48
Did
248k23225463
edited Jan 20 at 16:48
Did
248k23225463
edited Jan 20 at 16:48
Did
248k23225463
248k23225463
asked Jan 20 at 16:32
Yibei HeYibei He
3139
asked Jan 20 at 16:32
Yibei HeYibei He
3139
asked Jan 20 at 16:32
Yibei HeYibei He
3139
3139
$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53
add a comment |
$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53
$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53
add a comment |
1 Answer
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$begingroup$
$Y(x) > E[Y(x)]$ if and only if $X_i le x$ for all $i le E[Y(x)]$.
$$P(Y(x) > E[Y(x)]) = F(x)^{lfloorfrac{1}{1-F(x)}rfloor}.$$
[Your computation is close but you had a small typo. You should have $sum_{k ge lfloor frac{1}{1-F(x)}rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x to infty$ leads to the indeterminate form $1^infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as
$$frac{1}{1-F(x)} log F(x) le lfloor frac{1}{1-F(x)} rfloor log F(x) le (1+frac{1}{1-F(x)}) log F(x).$$
You can show that the left-hand and right-hand sides tend to a common limit as $x to infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
$endgroup$
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
add a comment |
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$begingroup$
$Y(x) > E[Y(x)]$ if and only if $X_i le x$ for all $i le E[Y(x)]$.
$$P(Y(x) > E[Y(x)]) = F(x)^{lfloorfrac{1}{1-F(x)}rfloor}.$$
[Your computation is close but you had a small typo. You should have $sum_{k ge lfloor frac{1}{1-F(x)}rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x to infty$ leads to the indeterminate form $1^infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as
$$frac{1}{1-F(x)} log F(x) le lfloor frac{1}{1-F(x)} rfloor log F(x) le (1+frac{1}{1-F(x)}) log F(x).$$
You can show that the left-hand and right-hand sides tend to a common limit as $x to infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
$endgroup$
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
add a comment |
$begingroup$
$Y(x) > E[Y(x)]$ if and only if $X_i le x$ for all $i le E[Y(x)]$.
$$P(Y(x) > E[Y(x)]) = F(x)^{lfloorfrac{1}{1-F(x)}rfloor}.$$
[Your computation is close but you had a small typo. You should have $sum_{k ge lfloor frac{1}{1-F(x)}rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x to infty$ leads to the indeterminate form $1^infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as
$$frac{1}{1-F(x)} log F(x) le lfloor frac{1}{1-F(x)} rfloor log F(x) le (1+frac{1}{1-F(x)}) log F(x).$$
You can show that the left-hand and right-hand sides tend to a common limit as $x to infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
$endgroup$
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
add a comment |
$begingroup$
$Y(x) > E[Y(x)]$ if and only if $X_i le x$ for all $i le E[Y(x)]$.
$$P(Y(x) > E[Y(x)]) = F(x)^{lfloorfrac{1}{1-F(x)}rfloor}.$$
[Your computation is close but you had a small typo. You should have $sum_{k ge lfloor frac{1}{1-F(x)}rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x to infty$ leads to the indeterminate form $1^infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as
$$frac{1}{1-F(x)} log F(x) le lfloor frac{1}{1-F(x)} rfloor log F(x) le (1+frac{1}{1-F(x)}) log F(x).$$
You can show that the left-hand and right-hand sides tend to a common limit as $x to infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
$endgroup$
$Y(x) > E[Y(x)]$ if and only if $X_i le x$ for all $i le E[Y(x)]$.
$$P(Y(x) > E[Y(x)]) = F(x)^{lfloorfrac{1}{1-F(x)}rfloor}.$$
[Your computation is close but you had a small typo. You should have $sum_{k ge lfloor frac{1}{1-F(x)}rfloor + 1} F(x)^{k-1} (1-F(x))$ which equals the above.]
Taking $x to infty$ leads to the indeterminate form $1^infty$, so you need to be more careful when taking the limit. Try taking a logarithm before applying the limit.
The logarithm can be sandwiched as
$$frac{1}{1-F(x)} log F(x) le lfloor frac{1}{1-F(x)} rfloor log F(x) le (1+frac{1}{1-F(x)}) log F(x).$$
You can show that the left-hand and right-hand sides tend to a common limit as $x to infty$ (e.g., using l'Hôpital's rule), so the middle quantity tends to this limit as well. Then exponentiate everything to conclude.
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
answered Jan 20 at 16:46
angryavianangryavian
41.9k23381
41.9k23381
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
add a comment |
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
$begingroup$
So I got the limit of the power is -1, and the final answer will be $e^-1$. Thank you!
$endgroup$
– Yibei He
Jan 20 at 20:32
add a comment |
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$begingroup$
Everything is fine until $$P(Y(x)gt E[Y(x)])=sum_{k=n+1}^infty kF^{k-1}(x)(1-F(x))$$ where $n$ denotes the integer part of $E(F(x))$, but after that, one simply cannot understand what you are doing. Next step: what is the value of $$sum_{k=n+1}^infty kt^{k-1}(1-t)$$ for $t$ in $(0,1)$?
$endgroup$
– Did
Jan 20 at 16:45
$begingroup$
I wonder if this problem is related to the secretary problem.
$endgroup$
– angryavian
Jan 20 at 16:53