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Intuition behind scaling random variable
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I'm trying to understand what exactly does it mean to scale random variable (let's say by K). I was watching one of the videos from Khan Academy where he showed scaling continuous probability distribution.
Check out this video
probability-distributions random-variables
asked Jul 23 at 7:00
omjego
113
add a comment |Â
up vote
1
down vote
favorite
I'm trying to understand what exactly does it mean to scale random variable (let's say by K). I was watching one of the videos from Khan Academy where he showed scaling continuous probability distribution.
Check out this video
probability-distributions random-variables
asked Jul 23 at 7:00
omjego
113
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to understand what exactly does it mean to scale random variable (let's say by K). I was watching one of the videos from Khan Academy where he showed scaling continuous probability distribution.
Check out this video
probability-distributions random-variables
asked Jul 23 at 7:00
omjego
113
I'm trying to understand what exactly does it mean to scale random variable (let's say by K). I was watching one of the videos from Khan Academy where he showed scaling continuous probability distribution.
Check out this video
probability-distributions random-variables
asked Jul 23 at 7:00
omjego
113
asked Jul 23 at 7:00
omjego
113
asked Jul 23 at 7:00
omjego
113
asked Jul 23 at 7:00
omjego
113
113
add a comment |Â
add a comment |Â
1 Answer
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When the random variable is described mathematically it may not make sense. If the random variable arises in some context, based on that context, one can make sense of scaling.
The intuition does not vary between continuous and discrete variables.
Let $X$ be the number of international passengers arriving at an airport on a day. This is a random variable. Possibly one immigration officer (counter) might service 500 international passengers. So for the govt the more important random variable would be $frac1500 X$.
One can similarly imagine situations in a call centre: if the number of calls received is a random variable, the number of employees needed for attending calls would be obtained by scaling it suitably.
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
When the random variable is described mathematically it may not make sense. If the random variable arises in some context, based on that context, one can make sense of scaling.
The intuition does not vary between continuous and discrete variables.
Let $X$ be the number of international passengers arriving at an airport on a day. This is a random variable. Possibly one immigration officer (counter) might service 500 international passengers. So for the govt the more important random variable would be $frac1500 X$.
One can similarly imagine situations in a call centre: if the number of calls received is a random variable, the number of employees needed for attending calls would be obtained by scaling it suitably.
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
add a comment |Â
up vote
2
down vote
When the random variable is described mathematically it may not make sense. If the random variable arises in some context, based on that context, one can make sense of scaling.
The intuition does not vary between continuous and discrete variables.
Let $X$ be the number of international passengers arriving at an airport on a day. This is a random variable. Possibly one immigration officer (counter) might service 500 international passengers. So for the govt the more important random variable would be $frac1500 X$.
One can similarly imagine situations in a call centre: if the number of calls received is a random variable, the number of employees needed for attending calls would be obtained by scaling it suitably.
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
add a comment |Â
up vote
2
down vote
up vote
2
down vote
When the random variable is described mathematically it may not make sense. If the random variable arises in some context, based on that context, one can make sense of scaling.
The intuition does not vary between continuous and discrete variables.
Let $X$ be the number of international passengers arriving at an airport on a day. This is a random variable. Possibly one immigration officer (counter) might service 500 international passengers. So for the govt the more important random variable would be $frac1500 X$.
One can similarly imagine situations in a call centre: if the number of calls received is a random variable, the number of employees needed for attending calls would be obtained by scaling it suitably.
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
When the random variable is described mathematically it may not make sense. If the random variable arises in some context, based on that context, one can make sense of scaling.
The intuition does not vary between continuous and discrete variables.
Let $X$ be the number of international passengers arriving at an airport on a day. This is a random variable. Possibly one immigration officer (counter) might service 500 international passengers. So for the govt the more important random variable would be $frac1500 X$.
One can similarly imagine situations in a call centre: if the number of calls received is a random variable, the number of employees needed for attending calls would be obtained by scaling it suitably.
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
answered Jul 23 at 7:30
P Vanchinathan
13.9k12035
13.9k12035
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
add a comment |Â
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
Got it. So the probability for 1/500 X will still be the same as X, right? This works perfectly fine in case of discrete variables as the area remains the same. But for continuous variables the outcome space is infinite and scaling will change the area under the curve so probabilities of each point must be decreased by some amount (this part of the video I didn't get)
– omjego
Jul 23 at 9:50
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
No, they are not same. X is X and 1/500 X is 1/500 X. It is the same as in change of variable one does in definite integral. If $y=kx$ when $x$ varies from $a$ to $b$ , $y$ will vary between $ka$ to $kb$.
– P Vanchinathan
Jul 23 at 9:58
add a comment |Â
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