Convergence in distribution of minimum of IID random variables
$begingroup$
I'm stuck on the following problem and could use a hint:
Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:
$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$
Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.
So far I've figured out that:
$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$
But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.
If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.
Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.
calculus probability convergence exponential-distribution
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
$endgroup$
add a comment |
$begingroup$
I'm stuck on the following problem and could use a hint:
Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:
$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$
Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.
So far I've figured out that:
$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$
But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.
If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.
Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.
calculus probability convergence exponential-distribution
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
$endgroup$
$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06
$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45
add a comment |
$begingroup$
I'm stuck on the following problem and could use a hint:
Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:
$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$
Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.
So far I've figured out that:
$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$
But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.
If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.
Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.
calculus probability convergence exponential-distribution
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
$endgroup$
I'm stuck on the following problem and could use a hint:
Let $Z_1,ldots,Z_n$ be IID random variables with density $f$. Suppose that $mathbb{P}(Z_i > 0) = 1$ and that $lambda = lim_{x to 0^{+} } f(x) > 0$. Let:
$$X_n = nminlbrace Z_1,ldots,Z_nrbrace$$
Show that $X_n rightsquigarrow Z$ (converges in distribution) where $Z$ has an exponential distribution with mean $frac{1}{lambda}$.
So far I've figured out that:
$$
begin{align*}
mathbb{P}(X_n leq x) &= mathbb{P}(nminlbrace Z_1,ldots,Z_nrbrace leq x) \
&= mathbb{P}(Z_1 leq x/n cup ldots cup Z_n leq x/n) \
&= 1 - mathbb{P}(Z_1 > x/n cap ldots cap Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n) cdot ldots cdot mathbb{P}(Z_n > x/n) \
&= 1 - mathbb{P}(Z_1 > x/n)^n \
&= 1 - (1-mathbb{P}(Z_1 leq x/n))^n \
&= 1 - (1-F_{Z_1}(x/n))^n
end{align*}
$$
But I'm not sure where to go from here. Although I know the CDF/PDF for $Z$, I don't know it for $Z_1,ldots,Z_n$, and it's not immediately obvious to me how I can leverage the fact that $Z_1,ldots,Z_n$ are guaranteed to be positive, other than if I rewrite $F_{Z_1}(x/n)$ to be an explicit integral of $f$ that the lower bound of the integral can be zero instead of negative infinity.
If I just try to take the limit as $n$ goes to infinity of both sides I can't simplify the RHS expression any further because of the power of $n$. I know that $a_n to a implies (1 + frac{a_n}{n})^n to e^a$ but I haven't managed to successfully apply it -- I assume this somehow helps me collapse the RHS into the CDF for an exponential, because I don't see another way to get $e$.
Finally I don't have any idea how to leverage the definition of lambda. Usually we care about right continuity of the CDF not the density function.
calculus probability convergence exponential-distribution
calculus probability convergence exponential-distribution
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
edited Dec 16 '18 at 22:40
Joseph Garvin
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
asked Dec 16 '18 at 22:22
Joseph GarvinJoseph Garvin
44928
44928
$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06
$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45
add a comment |
$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06
$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45
$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06
$begingroup$
@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
$endgroup$
– Joseph Garvin
Jan 8 at 23:06
$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45
$begingroup$
But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
$endgroup$
– Mike Hawk
Jan 9 at 16:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
|
show 1 more comment
$begingroup$
You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
|
show 1 more comment
$begingroup$
You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
$endgroup$
You need to prove this probability approximates $1-exp -lambda x$, i.e. that $(1-F_{Z_i}(x/n))^napproxexp -lambda x$. But for $ngg x$, $$F_{Z_i}(0)=0implies F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)=frac{lambda x}{n},$$so the claim follows from $exp -t=lim_{ntoinfty}(1-frac{t}{n})^n$.
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
answered Dec 16 '18 at 22:50
J.G.J.G.
26.3k22541
26.3k22541
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
|
show 1 more comment
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
$begingroup$
what does the n >> x notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 18:59
1
1
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
@JosephGarvin That $n$ is much greater than $x$.
$endgroup$
– J.G.
Jan 6 at 19:20
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
what does the $sim$ notation mean?
$endgroup$
– Joseph Garvin
Jan 6 at 21:54
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
@JosephGarvin is asymptotic to.
$endgroup$
– J.G.
Jan 6 at 21:55
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
$begingroup$
I don't understand how you conclude $ F_{Z_i}(x/n)sim frac{x}{n} F_{Z_i}'(0)$ ? how're those expressions related?
$endgroup$
– Joseph Garvin
Jan 9 at 0:00
|
show 1 more comment
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$begingroup$
Consider the Taylor series of $F_{Z_1}$ at x=0. Also, I suspect that you have gotten your X's and Z's mixed up some places.
$endgroup$
– Mike Hawk
Dec 16 '18 at 22:28
$begingroup$
@MikeHawk You were right about mixing up, copy and pasted a latex template and modified it and made a few mistakes. Hopefully fixed now...
$endgroup$
– Joseph Garvin
Dec 16 '18 at 22:42
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@MikeHawk: I don't see how I can use that Taylor series, because I don't know the CDF for $Z_1$, only for $Z$.
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– Joseph Garvin
Jan 8 at 23:06
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But you know $F_{Z_1}'(0)=lambda$, so you just have to show that $(1-lambda x/n+O(1/n^2))^nto e^{-lambda x}$
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– Mike Hawk
Jan 9 at 16:45