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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 ______________ .

1. Can someone please tell any good resources to learn about Gaussian Random variable?
2.  Is Gaussian random variable same as Normal distribution?

@jiminpark

1. JBStatistics videos on youtube is a good start.
2. yes, Normal distribution is also kmown as gaussian distribution.

@abcS Thank you Sir !

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

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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.

Funny name.
mean =0 and variance=1 represents standard normal distribution. How , come 1 equal to σ2 . As σ can be any value.
The point was very simple, nor a normal distribution, MEAN= MEDIAN. Hence 0

@ijnuhb
It’s be like mean=median=mode=0 in Standard Normal

$Y = max(0,X)$

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 $=0$

by

At top point, how r  u getting mean $\mu =0?$
given in question that mean=0
Are u saying

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

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.

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;

by

Is mode also zero here?
yes
For Gaussian distribution, Mean=MEDIAN=Mode, hence for the random variable X, Median = Mean=0. This also means that if we arrange all the real values taken by X with repetition in increasing order, then the sequence would look like

……….-3,-3,-2,-2,-2,-1,-1,-1,-1,-1,0,0,0,0,0(Median),0,0,0,0,1,1,1,1,1,2,2,2,3,3…...{Sequence is not exactly scaled according to the Gaussian Distribution, but it roughly shows how listing the whole distribution would look like. Also 0 being the mode, I have listed it comparatively more no. of times.}

Now, random variable Y=max(X,0)={0,for Y<=0

X,for Y>0}

So if we list distribution of Y, it would look like-

………..0,0,0,0,0,0,0,0,0,0,0,0,0,0,0(Median),0,0,0,0,1,1,1,1,1,2,2,2,3,3…….

since median is position dependent, so 0 remains the Median since only the entries smaller than it are replaced by 0 itself, without affecting its position. Also note that Mode of Y is 0 too. But mean of Y will shift away from 0 in the positive direction, as the distribution has become skewed.