Skip to main content
Cross Validated

Return to Question

added 2 characters in body
Source Link
Xi'an
Xi'an
  • 107.7k
  • 13
  • 190
  • 676

The question is

Let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$$f_X(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$$$F_Y(Y ≤ y) = P(2X-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

The question is

Let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

The question is

Let $X$ be a continuous random variable with pdf $f_X(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2X-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

Formatting
Source Link
edit approved Jan 18, 2021 at 7:22
MarianD
edit approved Jan 18, 2021 at 7:22
MarianD
  • 1.5k
  • 2
  • 11
  • 18

The question is "let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$."

Let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

The question is "let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$."

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

The question is

Let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

deleted 169 characters in body; edited tags
Source Link
Xi'an
Xi'an
  • 107.7k
  • 13
  • 190
  • 676

The question is "let X$X$ be a continuous random variable with pdf f(x) = 2(1 − x)$f(x) = 2(1 − x)$, 0 ≤ x ≤ 1$0 ≤ x ≤ 1$. If Y = 2X − 1$Y = 2X − 1$, find the pdf of Y$Y$."

I understand these steps: Fy(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = Fx((y+1)/2).$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of Y$Y$ from this. I know that we are supposed to differentiate both sides with respect to y$y$, but I do not understand what that means. I get that this is supposed to be really obvious, but I just do not understand how to do it. I have looked around a bunch, but I still am confused. Thank you so, so much for your help.

The question is "let X be a continuous random variable with pdf f(x) = 2(1 − x), 0 ≤ x ≤ 1. If Y = 2X − 1, find the pdf of Y."

I understand these steps: Fy(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = Fx((y+1)/2).

I do not understand how to get the pdf of Y from this. I know that we are supposed to differentiate both sides with respect to y, but I do not understand what that means. I get that this is supposed to be really obvious, but I just do not understand how to do it. I have looked around a bunch, but I still am confused. Thank you so, so much for your help.

The question is "let $X$ be a continuous random variable with pdf $f(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$."

I understand these steps$$F_Y(Y ≤ y) = P(2x-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$

I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.

Source Link
katie
katie
  • 23
  • 1
  • 4