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**Definition:** The mode of a sample or random variable is the value which occurs most often or with largest probability among all its values.
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<br>
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1) Let $x = \left\lbrace x_1, \ldots, x_n \right\rbrace$ be a [sample](/D/samp) from a [random variable](/D/rvar) $X$. Then, the mode of $x$ is the value which occurs most often in the list $x_1, \ldots, x_n$.
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<br>
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2) Let $X$ be a [random variable](/D/rvar) with [probability mass function](/D/pmf) or [probability density function](/D/pdf) $f_X(x)$. Then, the mode of $X$ is the the value which maximizes the PMF or PDF:
title: "Relation between gamma distribution and standard gamma distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Gamma distribution"
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theorem: "Relation to standard gamma distribution"
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sources:
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proof_id: "P177"
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shortcut: "gam-sgam2"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [gamma distribution](/D/gam) with shape $a$ and rate $b$:
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$$ \label{eq:X-gam}
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X \sim \mathrm{Gam}(a,b) \; .
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$$
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Then, the quantity $Y = b X$ will have a [standard gamma distribution](/D/sgam) with shape $a$ and rate $1$:
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$$ \label{eq:Y-snorm}
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Y = b X \sim \mathrm{Gam}(a,1) \; .
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$$
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**Proof:** Note that $Y$ is a function of $X$
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$$ \label{eq:Y-X}
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Y = g(X) = b X
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$$
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with the inverse function
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$$ \label{eq:X-Y}
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X = g^{-1}(Y) = \frac{1}{b} Y \; .
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$$
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Because $b$ is positive, $g(X)$ is strictly increasing and we can calculate the [probability density function of a strictly increasing function](/P/pdf-sifct) as
where $\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace$. With the [probability density function of the gamma distribution](/P/gam-pdf), we have
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