Finding the median of a distribution

[doctordiddy]

Homework Statement

Suppose X has the Uniform (0,1) distribution. Find the median of the distribution of e^X correct to 2 decimals.

Homework Equations

F(X) = 0.5
F(X) = ∫f(x)

The Attempt at a Solution

I am not entirely sure what to do here, I know to find the median you need to find when F(X) = 0.5. I assumed that e^X was just the cdf f(x) = e^X which makes F(x) still e^X.

So then I used e^(X) = 0.5 which gives X = ln(0.5).

However this is not correct. I am a bit confused about the first part of the question, how X being a uniform distribution would affect it.

Any help would be appreciated

Thanks

 

[Mark44]

Homework Statement

Suppose X has the Uniform (0,1) distribution. Find the median of the distribution of e^X correct to 2 decimals.

Homework Equations

F(X) = 0.5
F(X) = ∫f(x)

The Attempt at a Solution

I am not entirely sure what to do here, I know to find the median you need to find when F(X) = 0.5. I assumed that e^X was just the cdf f(x) = e^X which makes F(x) still e^X.

So then I used e^(X) = 0.5 which gives X = ln(0.5).

If you think about it a bit, this can't be right, since X would have to be a negative number. If X ranges from 0 to 1, then e^X will range from ? to ?

However this is not correct. I am a bit confused about the first part of the question, how X being a uniform distribution would affect it.

Any help would be appreciated

Thanks

 

[doctordiddy]

I found the answer to be e^0.5, but I am not sure exactly why. Do we find the median of the uniform distribution of X (0.5) and then plug that into e^(X)? Or do I need to find the median of the distribution between 1 and e? If it's the latter, how do we know that the median is simply e^(0.5)? Are we able to say this because X is a uniform distribution?

 

[doctordiddy]

If you think about it a bit, this can't be right, since X would have to be a negative number. If X ranges from 0 to 1, then e^X will range from ? to ?

I found the answer to be e^0.5, but I am not sure exactly why. Do we find the median of the uniform distribution of X (0.5) and then plug that into e^(X)? Or do I need to find the median of the distribution between 1 and e? If it's the latter, how do we know that the median is simply e^(0.5)? Are we able to say this because X is a uniform distribution?

 

[Ray Vickson]

I found the answer to be e^0.5, but I am not sure exactly why. Do we find the median of the uniform distribution of X (0.5) and then plug that into e^(X)? Or do I need to find the median of the distribution between 1 and e? If it's the latter, how do we know that the median is simply e^(0.5)? Are we able to say this because X is a uniform distribution?

If $Y = e^X$, the cumulative distribution of $Y$ is $F(y) = P(Y \leq y) = P(e^X \leq y)$. For $X \sim U(0,1)$, what does this give you for $F(y)$? What is the solution of $F(y) = 1/2$? That would give you the median.

 

[doctordiddy]

If $Y = e^X$, the cumulative distribution of $Y$ is $F(y) = P(Y \leq y) = P(e^X \leq y)$. For $X \sim U(0,1)$, what does this give you for $F(y)$? What is the solution of $F(y) = 1/2$? That would give you the median.

I am unsure of how to continue after
For $X \sim U(0,1)$, what does this give you for $F(y)$? Am I supposed to see that X = ln(y) or am I supposed to see that Y ~ U(1,e)? Or am I missing something altogether?

 

[haruspex]

Am I supposed to see that X = ln(y)

Some risk of confusion in the notation. That X = ln(Y) is self evident, so I'm assuming your question is why x' = ln(y'), x' and y' representing the medians of the respective distributions.
As Ray wrote, for any x, the event Y < e^x is the same event as X < x. Therefore F_Y(e^x) = F_X(x). Setting x=x' we get F_Y(e^x') = 0.5.

am I supposed to see that Y ~ U(1,e)

That is certainly not the case.