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Let X be a random variable having a Standard Uniform distribution. Let Y = √ X. Evaluate the mean and variance of Y
Original post by Yiuyu
Let X be a random variable having a Standard Uniform distribution. Let Y = √ X. Evaluate the mean and variance of Y


What have you tried?
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Yiuyu
OP
10
Original post by RDKGames
What have you tried?

Well for the mean I know that E[X]=1/2(b-a)=1/2 as X-U (0,1) however I need root x?
Original post by Yiuyu
Well for the mean I know that E[X]=1/2(b-a)=1/2 as X-U (0,1) however I need root x?


If you know E[X]\mathbb{E}[X] then you know E[Y2]\mathbb{E}[Y^2] which will be useful in finding var(Y)\mathrm{var}(Y)

Note that E[Y]=E[X]=∫abxf(x).dx\displaystyle \mathbb{E}[Y]=\mathbb{E}[\sqrt{X}]=\int_a^b \sqrt{x} f(x).dx which you can find.
(edited 6 years ago)
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Yiuyu
OP
10
Original post by RDKGames
If you know E[X]\mathbb{E}[X] then you know E[Y2]\mathbb{E}[Y^2] which will be useful in finding var(Y)\mathrm{var}(Y)

Note that E[Y]=E[X]=∫abxf(x).dx\displaystyle \mathbb{E}[Y]=\mathbb{E}[\sqrt{X}]=\int_a^b \sqrt{x} f(x).dx which you can find.


Okay worked out E[Y^2]=2/3 and need to find out Var(Y) which is E(X2) - [E(X)]^2 and Ik the second part so I have Var(Y) = something - 2/3^2 , how do i find that something or is it just 1/2( 1/1-0)
Original post by Yiuyu
Okay worked out E[Y^2]=2/3 and need to find out Var(Y) which is E(X2) - [E(X)]^2 and Ik the second part so I have Var(Y) = something - 2/3^2 , how do i find that something or is it just 1/2( 1/1-0)


Few errors going on there.

Firstly; E[Y2]≠23\mathbb{E}[Y^2] \neq \frac{2}{3}. In fact, you know that E[X]=12\mathbb{E}[X]=\frac{1}{2}, and since Y=XY=\sqrt{X}, we must have Y2=XY^2=X, hence E[X]=E[Y2]=12\mathbb{E}[X]=\mathbb{E}[Y^2]=\frac{1}{2}

Secondly; var(Y)\mathrm{var}(Y) is given by E[Y2]−E[Y]2\mathbb{E}[Y^2]-\mathbb{E}[Y]^2

Now you can finish it off.
Original post by RDKGames
Few errors going on there.

Firstly; E[Y2]≠23\mathbb{E}[Y^2] \neq \frac{2}{3}. In fact, you know that E[X]=12\mathbb{E}[X]=\frac{1}{2}, and since Y=XY=\sqrt{X}, we must have Y2=XY^2=X, hence E[X]=E[Y2]=12\mathbb{E}[X]=\mathbb{E}[Y^2]=\frac{1}{2}

Secondly; var(Y)\mathrm{var}(Y) is given by E[Y2]−E[Y]2\mathbb{E}[Y^2]-\mathbb{E}[Y]^2

Now you can finish it off.


Original post by Yiuyu
Okay worked out E[Y^2]=2/3 and need to find out Var(Y) which is E(X2) - [E(X)]^2 and Ik the second part so I have Var(Y) = something - 2/3^2 , how do i find that something or is it just 1/2( 1/1-0)


Would something like this actually come up in an S1 test? :O
Original post by xxxtentacion..
Would something like this actually come up in an S1 test? :O


Depends on what S1 he is talking about. He could be doing some odd board as an A-Level, or it could be a uni module on stats.

Just look in your own spec to see what you'd encounter.

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