Finding density function of continuous random variable
$begingroup$
A stick of length L is broken at a uniformly chosen random point. Let
$X$ be the length of the smaller piece. Find the density of $X$.
try
Notice that the probability that the stick is broken a certain is equally likely. So, we have that smaller piece is either $[0,x]$ or $[x,L]$
$$ P(X leq x) = frac{ text{length from origin to x}+text{length from x to L}}{text{total length} }= frac{x + |x-L|}{L}$$
But, this leads anywhere. Instead I was thinking that the only way the $X$ is the smaller piece is if $X$ is less than $L/2$ thus $X$ is uniform on $L/2$ and s o
$$ f_X(x) = frac{2}{L} $$
Is this correct?
probability
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
$endgroup$
add a comment |
$begingroup$
A stick of length L is broken at a uniformly chosen random point. Let
$X$ be the length of the smaller piece. Find the density of $X$.
try
Notice that the probability that the stick is broken a certain is equally likely. So, we have that smaller piece is either $[0,x]$ or $[x,L]$
$$ P(X leq x) = frac{ text{length from origin to x}+text{length from x to L}}{text{total length} }= frac{x + |x-L|}{L}$$
But, this leads anywhere. Instead I was thinking that the only way the $X$ is the smaller piece is if $X$ is less than $L/2$ thus $X$ is uniform on $L/2$ and s o
$$ f_X(x) = frac{2}{L} $$
Is this correct?
probability
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
$endgroup$
$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37
add a comment |
$begingroup$
A stick of length L is broken at a uniformly chosen random point. Let
$X$ be the length of the smaller piece. Find the density of $X$.
try
Notice that the probability that the stick is broken a certain is equally likely. So, we have that smaller piece is either $[0,x]$ or $[x,L]$
$$ P(X leq x) = frac{ text{length from origin to x}+text{length from x to L}}{text{total length} }= frac{x + |x-L|}{L}$$
But, this leads anywhere. Instead I was thinking that the only way the $X$ is the smaller piece is if $X$ is less than $L/2$ thus $X$ is uniform on $L/2$ and s o
$$ f_X(x) = frac{2}{L} $$
Is this correct?
probability
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
$endgroup$
A stick of length L is broken at a uniformly chosen random point. Let
$X$ be the length of the smaller piece. Find the density of $X$.
try
Notice that the probability that the stick is broken a certain is equally likely. So, we have that smaller piece is either $[0,x]$ or $[x,L]$
$$ P(X leq x) = frac{ text{length from origin to x}+text{length from x to L}}{text{total length} }= frac{x + |x-L|}{L}$$
But, this leads anywhere. Instead I was thinking that the only way the $X$ is the smaller piece is if $X$ is less than $L/2$ thus $X$ is uniform on $L/2$ and s o
$$ f_X(x) = frac{2}{L} $$
Is this correct?
probability
probability
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
asked Dec 15 '18 at 4:50
Jimmy SabaterJimmy Sabater
2,146219
2,146219
$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37
add a comment |
$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37
$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37
add a comment |
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$begingroup$
$Y sim U(0,L)$ Find $Y=min(X,L-X)$
$endgroup$
– Daman deep
Dec 15 '18 at 5:03
$begingroup$
"But, this leads anywhere" This very much leads to the result, only one should be more careful: for $x$ in $(0,L/2)$, the event ${X<x}$ corresponds to the location of the break being in $[0,x]$ or in $[L-x,L]$, thus, $$P(X<x)=frac{x+(L-(L-x))}L=frac{2x}L$$ and you are done.
$endgroup$
– Did
Dec 15 '18 at 6:37