Suppose I have two Beta random variables $$ X \sim \text{Beta}(a, b), \\ Y \sim \text{Beta}(ca, cb), $$ with $c > 1$. They both have same mean $\tfrac{a}{a+b}$, but $Y$ is more "peaky" than $X$. My question is as follows. Let $t \ge 0$; does $$ \mathbb{E}(e^{-tX}) \ge \mathbb{E}(e^{-kY}) \;\; \text{?} $$ My intuition says yes, because $X$ "spreads" the weight more, therefore taps into the places where $e^{-kx}$ is large. The case $t = 0$ is trivial, but I would like to prove it for $t > 0$.

Here is a mathematica plot, with $t = 4$, and $(a, b) \in \{(2,4), (4,8), (8,16)\}$.

p1 = Plot[
   Evaluate@
    Table[PDF[BetaDistribution[\[Alpha], 2*\[Alpha]], 
      x], {\[Alpha], {2, 4, 8}}], {x, 0, 1}, Filling -> Axis];
p2 = Plot[Exp[-4 x], {x, 0, 1}];
Show[p1, p2]

and the expectations, providing some empirical support to my claim:

Plot[Hypergeometric1F1[x, 3 x, -4], {x, 2, 8}, Frame -> True]

Note: I know that the expectation I'm trying to compute is just the moment generating function of the Beta distribution evaluated at negative values. This MGF is Kummer's confluent hypergeometric function of the first kind: $\mathbb{E}(e^{-tX}) = {}_1F_1(a, a+b, -t)$. I've spent quite a bit of time investigating this function, and nothing came out of it.

Suppose I have two Beta random variables $$ X \sim \text{Beta}(a, b), \\ Y \sim \text{Beta}(ca, cb), $$ with $c > 1$. They both have same mean $\tfrac{a}{a+b}$, but $Y$ is more "peaky" than $X$. My question is as follows. Let $t \ge 0$; does $$ \mathbb{E}(e^{-tX}) \ge \mathbb{E}(e^{-tY}) \;\; \text{?} $$ My intuition says yes, because $X$ "spreads" the weight more, therefore taps into the places where $e^{-tx}$ is large. The case $t = 0$ is trivial, but I would like to prove it for $t > 0$.

Here is a mathematica plot, with $t = 4$, and $(a, b) \in \{(2,4), (4,8), (8,16)\}$.

p1 = Plot[
   Evaluate@
    Table[PDF[BetaDistribution[\[Alpha], 2*\[Alpha]], 
      x], {\[Alpha], {2, 4, 8}}], {x, 0, 1}, Filling -> Axis];
p2 = Plot[Exp[-4 x], {x, 0, 1}];
Show[p1, p2]

and the expectations, providing some empirical support to my claim:

Plot[Hypergeometric1F1[x, 3 x, -4], {x, 2, 8}, Frame -> True]

Note: I know that the expectation I'm trying to compute is just the moment generating function of the Beta distribution evaluated at negative values. This MGF is Kummer's confluent hypergeometric function of the first kind: $\mathbb{E}(e^{-tX}) = {}_1F_1(a, a+b, -t)$. I've spent quite a bit of time investigating this function, and nothing came out of it.