GATE CSE 2017 | Set 1 | Question: 19

Question

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 ______________ .

Comments

Ans

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Y = max(X,0) gives Y=0 for all X<0 (-ve values) 

Y = max(X,0) gives Y=0 for X =0

Y = max(X,0) gives Y=X for all X>0 (all natural numbers) 

a gaussian/ normal distribution is symmetric around the mean 

that means , for a range of  X like [-5 ,5 ] , we get output of Y = max (X,0) as 

0,0,0,0,0,0,1,2,3,4,5 ---> now find the median , it is 0

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Since already 50% probability lies at 0, the median is 0 ?

Answer 1

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.

Comments

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

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Hoping that I’m not mistaken, I think \(0\) shouldn't be the median of \(Y\) (at least by the arguments of the above accepted answer). A short proof to support my claim is as follows:-

Suppose, for the sake of contradiction, \(c=0\) is indeed the median of \(Y\). Plugging the value of \(c\) in

\[P\{0<Y<c\}=P\{c<Y<\infty\},\]

we get, \[P\{0<Y<0\}=P\{0<Y<\infty\},\] which is impossible, as no real number can be both less than and greater than \(0\) (or any other real number for that matter) at the same time. This implies that either the above definition of the median of a probability distribution is wrong, or \(0\) is not the median of \(Y\).

Suppose it’s the first case, and we relax our definition of the median of a probability distribution to

\[P\{0\leq Y\leq c\}=P\{c\leq Y<\infty\}.\]

Then putting the value of \(c=0\) in the above equation, we get,

\[P\{0\leq Y\leq 0\}=P\{0\leq Y<\infty\},\]

\[or,\ P\{Y=0\}=P\{0\leq Y<\infty\},\]

\[or,\ 0=P\{0\leq Y<\infty\}.\]

\(\Bigr[\)The last equation above follows from the second last equation above it because \(Y\) is a continuous random variable (as \(X\) is a continuous random variable), and for any continuous random variable \(Y\), with a probability density function \(f(x)\) and cumulative distribution function \(F(x)\), we have,

\[P\{Y=a, \text{where} \ a \in \mathbb{R}\} = \int_{a}^{a} f(x) \, dx = F(x) \Bigr|_{a}^{a} = F(a)-F(a)=0.\Bigr]\]

But, we know that \[0\neq P\{0\leq Y<\infty\},\] which is a contradiction, hence proving the claim that \(0\) is not the median of \(Y\).

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@John_Smith you were half-right here, but Y is not a continous random variable, thus at Y=0, its probablity is not 0, IT is 1/2 and thats equal to RHS.

ref: https://en.wikipedia.org/wiki/Random_variable#Mixed_type

Answer 2

$Y = max(0,X)$

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 $=0$

Comments

At top point, how r  u getting mean $\mu =0?$

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given in question that mean=0

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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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Brilliant answer

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Great explanation

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On point explanation 💯

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Beautifully Explained!

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understood thanks!🙂

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excellent expalination

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Best Explanation

Answer 3

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.

Answer 4

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;

Comments

Is mode also zero here?

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yes

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Yes , because for all the negative values of X the function Y becomes 0 . So 0 is mostly repeated

Answer 5

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.

Comments

How mean of y will not be zero

Answer 6

Ans