Let $Y = max(X, 0)$, then median of $Y$ is ____

Let $Y = max(X, 0)$, then median of $Y$ is _____.

Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.

My try:

Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.

Can you please explain in other way?

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

add a comment |

Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.

My try:

Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.

Can you please explain in other way?

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

add a comment |

Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.

My try:

Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.

Can you please explain in other way?

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

Let $X$ be a Gaussian random variable with mean $0$ and variance $σ^2$. Let $Y = max(X, 0)$ where $max(a, b)$ is the maximum of $a$ and $b$. The median of $Y$ is _____.

My try:

Somewhere it explain as: Here, half of the values of Y are to the left of the mean X = 0 and the remaining half of the values of Y lies to the right of the mean X = 0. hence,The median of Y = 0.

Can you please explain in other way?

probability probability-distributions random-variables means median

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

asked Sep 22 '17 at 10:50

Mithlesh Upadhyay

2,90982864

add a comment |

3 Answers
3

active

oldest

votes

You have for $X$ a continuous random variable with median $0$

$P(X le 0)=frac12$

$P(X gt 0)=frac12$

so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,

$P(Y =0)=frac12$

$P(Y gt 0)=frac12$

so, moving towards the definition of a median,

$P(Y le 0) ge frac12$

$P(Y ge 0) ge frac12$

so the median of $Y$ is $0$

answered Sep 22 '17 at 11:17

Henry

98.4k475162

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

add a comment |

The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.

Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$

... since $X$ is symmetrically distributed about $0$.

Thus $0$ is the median of $Y$.

That is all there is too it.

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

add a comment |

-1

As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0

answered Nov 28 '18 at 16:18

Geeklovenerds

529

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You have for $X$ a continuous random variable with median $0$

$P(X le 0)=frac12$

$P(X gt 0)=frac12$

so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,

$P(Y =0)=frac12$

$P(Y gt 0)=frac12$

so, moving towards the definition of a median,

$P(Y le 0) ge frac12$

$P(Y ge 0) ge frac12$

so the median of $Y$ is $0$

answered Sep 22 '17 at 11:17

Henry

98.4k475162

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

add a comment |

You have for $X$ a continuous random variable with median $0$

$P(X le 0)=frac12$

$P(X gt 0)=frac12$

so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,

$P(Y =0)=frac12$

$P(Y gt 0)=frac12$

so, moving towards the definition of a median,

$P(Y le 0) ge frac12$

$P(Y ge 0) ge frac12$

so the median of $Y$ is $0$

answered Sep 22 '17 at 11:17

Henry

98.4k475162

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

add a comment |

You have for $X$ a continuous random variable with median $0$

$P(X le 0)=frac12$

$P(X gt 0)=frac12$

so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,

$P(Y =0)=frac12$

$P(Y gt 0)=frac12$

so, moving towards the definition of a median,

$P(Y le 0) ge frac12$

$P(Y ge 0) ge frac12$

so the median of $Y$ is $0$

answered Sep 22 '17 at 11:17

Henry

98.4k475162

You have for $X$ a continuous random variable with median $0$

$P(X le 0)=frac12$

$P(X gt 0)=frac12$

so, since $X le 0 implies Y=0$ and $Xgt 0 implies Y gt 0$,

$P(Y =0)=frac12$

$P(Y gt 0)=frac12$

so, moving towards the definition of a median,

$P(Y le 0) ge frac12$

$P(Y ge 0) ge frac12$

so the median of $Y$ is $0$

answered Sep 22 '17 at 11:17

Henry

98.4k475162

answered Sep 22 '17 at 11:17

Henry

98.4k475162

answered Sep 22 '17 at 11:17

Henry

98.4k475162

answered Sep 22 '17 at 11:17

Henry

98.4k475162

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

add a comment |

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

@Henry-What could be the density function of Y in this case?
– user3767495
Jul 10 '18 at 12:27

@user3767495 $Y$ will have the same density as $X$ for positive values, plus a point of positive probability $frac12$ (not density) at $0$
– Henry
Jul 10 '18 at 13:00

add a comment |

The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.

Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$

... since $X$ is symmetrically distributed about $0$.

Thus $0$ is the median of $Y$.

That is all there is too it.

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

add a comment |

The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.

Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$

... since $X$ is symmetrically distributed about $0$.

Thus $0$ is the median of $Y$.

That is all there is too it.

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

add a comment |

The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.

Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$

... since $X$ is symmetrically distributed about $0$.

Thus $0$ is the median of $Y$.

That is all there is too it.

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

The median of $Y$ is defined as the least value $m$ such that $mathsf P(Yleqslant m)geqslanttfrac 12$.

Since $Y=max{0,X}$, then: $mathsf P(Yleqslant 0){~=~mathsf P(Y=0)\~=~ mathsf P(Xleqslant 0)\~=~tfrac 12}$

... since $X$ is symmetrically distributed about $0$.

Thus $0$ is the median of $Y$.

That is all there is too it.

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

edited Sep 22 '17 at 11:23

community wiki

2 revs
Graham Kemp

community wiki

2 revs
Graham Kemp

community wiki

2 revs
Graham Kemp

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

add a comment |

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

Thanks for nice explanation.
– Mithlesh Upadhyay
Sep 22 '17 at 11:24

add a comment |

-1

As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0

answered Nov 28 '18 at 16:18

Geeklovenerds

529

add a comment |

-1

As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0

answered Nov 28 '18 at 16:18

Geeklovenerds

529

add a comment |

-1

As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0

answered Nov 28 '18 at 16:18

Geeklovenerds

529

As mean =0 and variance σ2 represents the standard normal distribution.And the graph of Standard normal distribution is symmetric And in symmetric we have,
Mean=Median=Mode.

As given that mean is 0.

Hence Median and mode will be 0

answered Nov 28 '18 at 16:18

Geeklovenerds

529

answered Nov 28 '18 at 16:18

Geeklovenerds

529

answered Nov 28 '18 at 16:18

Geeklovenerds

529

answered Nov 28 '18 at 16:18

Geeklovenerds

529

add a comment |

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