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Let $X$ be a Gaussian random variable with mean 0 and variance $\sigma ^{2}$. Let $Y$ = $\max\left ( X,0 \right )$ where $\max\left ( a,b \right )$ is the maximum of $a$ and $b$. The median of $Y$ is ______________ .
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@jiminpark 

  1. JBStatistics videos on youtube is a good start.
  2. yes, Normal distribution is also kmown as gaussian distribution.
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@abcS Thank you Sir !

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@jiminpark I’m not a sir. An aspirant, just like you. :)

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5 Answers

43 votes
43 votes
Best answer
Variable $Y$ can take only non-negative values. Median of a distribution is a value $c$ such that
$$P(0<Y<c) = P(c<Y<\infty)$$
Now for L.H.S., $Y$ will lie between $0$ and $c$ only when $X<c$ i.e $P(0<Y<c) = P(X<c)$.

For R.H.S,  $Y>c$ only when $X>c$ i.e. $P(c<Y<\infty) = P(X>c) = 1-P(X<c)$

Equating both sides, we get $P(X<c) = 1-P(X<c) \implies P(X<c) = 0.5 \implies c = 0$.

Hence $0$ is the answer.
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useless
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@swami_9 Are you trying to prove that you are useless?

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@Arjun this user @swami_9 is doing only these kind of things only...

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43 votes
43 votes

$Y = max(0,X)$

so value generated by $Y$ would depend on value of $X$

$case\ 1 :$ when $X<0$ then $Y = max(0, -ve)=0$

$case\ 2 :$ when $X =0$ then $Y =max(0,0)=0$

$case\ 3 :$ When $X >0$ then $Y = max(0,X)=X$.

 

So what would be the sequence of values generated for $Y$ ?

$....... \underbrace{0,0,0,0,0,0,0}_{\text{when X is negative}}\ 0\ \underbrace{1,2,3,4,5,6,7}_{\text{when X is positive}} ......$

 

This is because $X$ is a Gaussian random variable  $\implies$ its curve would be same as that of normal distribution.

We know curve of normal distribution is always bell shaped and is split into $2$ equal halves by the mean value.

and in question it is given that $mean = 0$ so half the values of $X$ would be $-ve$ and half of them would be $+ve$

For every $-ve$ value of $X$ , $Y=0$ and for every $+ve$ value of  $X$, $Y=X$

Now what is median ?

The middle value of among a sequence is called median , like median of $0,2,2, 3,9$ is $2$ because $2$ is at the middle.


Similarly for this sequence

$\underbrace{0,0,0,0,0,0,0}_{\text{when X is negative}}\ 0\ \underbrace{1,2,3,4,5,6,7}_{\text{when X is positive}}$

the median would be $0$

 

hence median of $Y$ = middle value among the sequence $=0$

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4 Comments

Are u saying

as mean is $0,$ median also be $0.$ As here median is depend on mean. right?
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Yes you can say that because in symmetric distributions , mean=median=mode.
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Clear explanation.
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10 votes
10 votes

Simply , It can be solved in following ways :-

We have for X a continuous random variable with median 0, that means :

  • P(X ≤ 0) = 1/2
  • P(X > 0) = 1/2

so, since X ≤ 0 → Y = 0 and X > 0 → Y > 0,

  • P(Y = 0) = 1/2
  • P(Y > 0) = 1/2

And definition of a median (m) of random variable Y is :

  • P(Y ≤ m) ≥ 1/2
  • P(Y ≥ m) ≥ 1/2

Therefore, median (m) of random variable Y is 0.

3 votes
3 votes

We can solve this by Simple Reasoning, that Median is Either X or 0(Zero if X is negative) and it's given that X is Gaussian Random Variable, if a Random Variable is Gaussian then it's Distribution is Gaussian/Normal Distribution.

So it's Normal Distribution and It's also given that Mean is 0, when mean is 0, Median is 0 always. you can also check the plot for different values of Mean and SD using https://www.desmos.com/calculator/0x3rpqtgrx

therefore Median is 0;

2 Comments

Is mode also zero here?
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yes
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Answer:

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