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$X \sim N (1,4) ;\, Y= 2X+1;$

And now does $Y$ have $N(1\times2+1,\, 2\times4+1) = N(3,9)$ ?

Is it valid? And if so why ? Or how do I get y's distribution?

Because when I run the following in R, it yields

x = dnorm(1,1,4) # [1] 0.09973557

y = dnorm(3,3,9) # [1] 0.04432692
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  • $\begingroup$ $Var(a \cdot X + b) = a^2 \cdot Var(x)$ $\endgroup$
    – Repmat
    Feb 16 '18 at 11:44
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Recall that $Var(aX+b) = a^2Var(X)$. So your distribution for $Y$ is almost correct, apart from the variance.

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The linear transformation of variables is shown here.

Suppose a linear transformation is applied to the random variable $X$ to create a new random variable $Y$. Then, the mean and variance of the new random variable $Y$ are defined by the following equations.

$$\bar{Y} = m\bar{X} + b \text{ and } Var(Y) = m^2 Var(X)$$

where $m$ and $b$ are constants, $\bar{Y}$ is the mean of $Y$, $\bar{X}$ is the mean of $X$, $Var(Y)$ is the variance of $Y$, and $Var(X)$ is the variance of $X$.

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    $\begingroup$ @atp: If an answer is helpful, please upvote and/or accept the answer. $\endgroup$
    – user64106
    Feb 16 '18 at 15:15
  • $\begingroup$ @StatMan (+1) For your answer, I don't think OP understood. $\endgroup$
    – Carl
    Feb 16 '18 at 18:01

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