Confusion in limit
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So, I was going through a proof about stochastic convergence in the book written by Hogg & Craig.
At one point of the proof they write
$$1=lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty} F_n[(c+epsilon)-]-lim_{nto infty} F_n(c-epsilon).$$
I have a problem here.
$$lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty}[F_n[(c+epsilon)-]-F_n(c-epsilon)].$$
But how can they insert the limit an put it in front of the two functions? For doing that we have to know that both the limits have to exist. Am i missing something?
Any help will be appreciated!
calculus probability-theory
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
$endgroup$
add a comment |
$begingroup$
So, I was going through a proof about stochastic convergence in the book written by Hogg & Craig.
At one point of the proof they write
$$1=lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty} F_n[(c+epsilon)-]-lim_{nto infty} F_n(c-epsilon).$$
I have a problem here.
$$lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty}[F_n[(c+epsilon)-]-F_n(c-epsilon)].$$
But how can they insert the limit an put it in front of the two functions? For doing that we have to know that both the limits have to exist. Am i missing something?
Any help will be appreciated!
calculus probability-theory
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
$endgroup$
$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25
add a comment |
$begingroup$
So, I was going through a proof about stochastic convergence in the book written by Hogg & Craig.
At one point of the proof they write
$$1=lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty} F_n[(c+epsilon)-]-lim_{nto infty} F_n(c-epsilon).$$
I have a problem here.
$$lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty}[F_n[(c+epsilon)-]-F_n(c-epsilon)].$$
But how can they insert the limit an put it in front of the two functions? For doing that we have to know that both the limits have to exist. Am i missing something?
Any help will be appreciated!
calculus probability-theory
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
$endgroup$
So, I was going through a proof about stochastic convergence in the book written by Hogg & Craig.
At one point of the proof they write
$$1=lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty} F_n[(c+epsilon)-]-lim_{nto infty} F_n(c-epsilon).$$
I have a problem here.
$$lim_{nto infty} Pr(|Y_n -c|< epsilon)=lim_{nto infty}[F_n[(c+epsilon)-]-F_n(c-epsilon)].$$
But how can they insert the limit an put it in front of the two functions? For doing that we have to know that both the limits have to exist. Am i missing something?
Any help will be appreciated!
calculus probability-theory
calculus probability-theory
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
edited Jan 11 at 5:41
Lord Shark the Unknown
108k1162136
108k1162136
asked Jan 11 at 3:49
user587126user587126
255
asked Jan 11 at 3:49
user587126user587126
255
asked Jan 11 at 3:49
user587126user587126
255
255
$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25
add a comment |
$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25
$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25
$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25
add a comment |
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$begingroup$
You are right in saying that individual terms need not have a limit (in general) but it is difficult to give a definite answer unless you give more details on what Hogg and Craig are doing.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 6:25